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PLANE AND SPHERICAL 
TRIGONOMETRY 



BY 

JAMES M. TAYLOR 



TEACHERS' EDITION 



GINN & COMPANY 

BOSTON • NEW YORK • CHICAGO ■ LONDON 



LIBRARY of CONGRESS 
Two Copies Received 

AUG 16 1906 

/i Copy riji u\ Entry 
fLASS CK. XXc. No, 
COPY B, 



Jilt 



COPYRIGHT, 1906 
BY JAMES M. TAYLOR 



ALL EIGHTS RESERVED 
66.8 



GINN & COMPANY- PRO- 
PRIETORS • BOSTON • U.S.A. 



PREFACE 

This book contains the solutions of all the examples in 
Taylor's Plane and Spherical Trigonometry. Great care has 
been taken to make the solutions models in form and method, 
so that young teachers, by a judiciqns use of the book, can 
gain valuable aid and suggestions in methods of solution and 
systematic arrangement of work. This edition is intended 
for teachers, and for them only. 

The publishers will not sell the book except to teachers 
of Taylor's Trigonometry ; and every teacher must consider 
himself in honor bound not to sell his copy except to the 
publishers, Messrs. Ginn & Company. 

JAMES M. TAYLOR 



PLAXE TRIGONOMETRY 



EXERCISE I 



Construct the acute angle A, 
obtain the values of all its trigo- 
nometric ratios, and find its size 
in degrees : 

1. When sin A = 2/5. 

Here MP/ OP = 2/5. 

Hence if OP = 5, 2IP = 2. 

Draw OY± OX at 0, on OY lay- 
off OD = 2, and through D draw 
DN II to OX. With as a center 
and 5 as a radius draw an arc cut- 
ting DN in some point as P. Draw 
PM ± OX and draw OP ; then 
MP = 2, OP = 5; hence 

sin XOP = 2/5 = sin A. 
.-. Z XOP = A. 

031 = V5 2 - 2^ = V21. 
F 

\£ 

W 




Fig. 1 



MX 



sin J. = 2/5, esc A = 5/2 ; 
cos A — V21/5, sec A = 5/V21 ; 
tan A = 2/ V21, cot J. = V21/2. 
J. = 23° 34' 40". 



2. When sin A = 4/5. 
Here MP/ OP = 4/5. 
Hence if OP = 5, J/P = 4. 
Draw 0Y± OX at 0, on OY lay 

off OP = 4, and through P draw 
DN II to OX. With as a center 
and 5 as a radius draw an arc cut- 
ting DN in some point as P. Draw 
PM ± OX and draw OP ; then 
MP = 4, OP = 5 ; hence 

sin XOP = 4/5 = sin^. 
.-. ZX0P = A. 

OM = V52 - 4 2 = 3. 

.-. sin A = 4/5, csc^l = 5/4 ; 
cos A = 3/5, sec A = 5/3 ; 
tan A = 4/3, cot A = 3/4. 
J. = 53° 8'. 

3. When cos A = 3/4. 
Here OM/OP = 3/4. 
Hence if OP = 4, CLV = 3. 

On OX lay off 03/ = 3, and at 
M draw JP8 _L OX. With as a 
center and 4 as a radius draw an 
arc cutting MS in some point as P. 
Draw OP ; then 

cos XOP = 3/4 = cos^4. 

.-. ZX0P = A. 

MP = V42 _ 32 = V7. 



PLANE TRIGONOMETRY 



.-. sin A = V7/4, esc A = 4/V7; 
cos A = 3/4, sec A = 4/3 ; 
tan^.=V7A cot^l = 3/V7. 
^L = 41° 24' 30". 

4. When cos^L = 1/3. 

Here OM/OP = 1/3. 
Hence if OP = 3, OM = 1. 

On OX lay off Olf = 1, and at 
itf draw .MS JL OX. With as a 
center and 3 as a radius draw an 
arc cutting MS in some point as P. 
Draw OP ; then 

cos XOP = 1/3 = cos J.. 
.-. Z XOP = A. 

MP = V3 2 - .P = V8. 

.;. sin A = V8/3, esc ^4 = 3/V8 ; 
cosJ. = l/3, sec .1 = 3; 
tan A = V8, cot ^4 = 1/ V8. 
A = 70° 32'. 

5. When tan A = 1/4. 

Here MP/OM = 1/4. 
Hence if OJf = 4, JfP = I. 

On OX lay off OJf =4, and at 
M draw .MS ± OX. On MS lay off 
JfP = 1, and draw OP ; then 

tan XOP = 1/4 = tan .4. 

.-. ZZOP = i. 

V17. 



OP =V42+1 2 



sin ^1 = 1/V17, esc A — VI 7 ; 
cos J. = 4/V17, sec A = VI 7/4 
tan A = 1/4, cot A = 4. 
^1 = 14° 2' 15". 

6. When tan A = 4/3. 

Here MP/OM = 4/3. 
Hence if OAf = 3, MP = 4. 



On OX lay off OJf = 3, and at 
M draw M S _L OX. On MS lay off 
ifP = 4, and draw OP ; then 

tan XOP = 4/3= Ian A. 
.: ZXOP = A. 

OP = V3 2 + 42 = 5. 

.-. sin A = 4/5, esc ^4 = 5/4 ; 
cos A = 3/5, sec A = 5/3 ; 
tanJ. = 4/3, cotJ. = 3/4. 
A = 53° V 45". 

7. When cot .4 = 5/2. 

Here OM/MP = 5/2. 

Hence if MP = 2, OM = 5. 

On OX lay off OJW = 5, and at 
M draw MS ± OX. On MS lay off 
ifP = 2, and draw OP ; then 

cot XOP = 5/2 = cot.l. 

.-. Z XOP = A. 

OP = V52 + 22 = V29. 

.-. sin A = 2/ V29, esc A = V29/2 ; 
cos A = 5/ V29, sec A = V29/5 ; 
tan A = 2/5, cot A = 5/2. 
A = 21° 48' 6". 

8. When cotJ_ = 1/3. 

Here OM/MP = 1/3. 

Hence if M P = 3, OJf =1. 

On OX lay off Oilf = 1, and at 
M draw .MS _L OX. On .MS lay off 
MP = 3, and draw OP ; then 

cot XOP = 1/3 = cot^4. 
.-. Z XOP = A. 

op = V32 + 12 = vio. 

.-. sin A = 3/ V10, esc A = V10/3 ; 

cos^4 = l/V10, sec A= V10; 

tan^i = 3, cot^4 = l/3. 

A = 71° 34'. 



EXERCISE I 



3 



9. When sec A = 5/3. 

Here OP/OM = 5/3. 

Hence if OM = 3, OP = 5. 

On OX lay off OM = 3, and at 
M draw MS ± OX. With as a 
center and 5 as a radius, draw an 
arc cutting MS in some point as P. 
Draw OP ; then 

sec XOP = 5/3 = sec A . 
.: ZX0P = ^4. 

MP = V5 2 - 3 2 = 4. 

.-. sin ^4 = 4/5, esc A = 5/4 ; 
cos A = 3/5, sec A = 5/3 ; 
tanJ. = 4/3, cotJ. = 3/4. 
A = 53° 8'. 

10. When sec ^4 = 4/3. 

Here OP/ OM = 4/3. 
Hence if Oif = 3, OP = 4. 

On OX lay off OM = 3, and at 
M draw JtfS _L OX. With as a 
center and 4 as a radius draw an 
arc cutting MS in some point as P. 
Draw OP ; then 

sec XOP = 4/3 = sec^4. 
.-. Z XOP = A. 

MP = V42 - 3 2 = V7. 

.-. sin ^4 = V7/4, esc J. = 4/V7; 
cos ^4 = 3/4, sec A = 4/3 ; 
tan A = V7/3, cot J. = 3/V7. 
J. = 41° 24' 30". 

11. When esc A = 5/2. 

Here OP /MP = 5/2, 

Hence if MP = 2, OP = 5. 

Draw OF± OX at 0, on OF lay 

off OD = 2, and through D draw 

DiV II OX. With as a center and 

5 as a radius draw an arc cutting 



DN in some point as P. Draw 
PM ± OX, and draw OP ; then 
esc XOP = 5/2 = esc A. 
.-. ZX0P = A. 

OM = V52- 22= V21. 
.-. sin ^4 = 2/5, esc J. = 5/2 ; 
cos A = V21/5, sec A = 5/V21 ; 
tan A = 2/V21, cot ^4 = V21/2. 
^4 = 23° 35'. 

12. When cscJ. = 3/2. 
Here OP /MP = 3/2. 
Hence if ifP = 2, OP = 3. 
Draw 0Y± OX at 0, on OY lay 

off OD = 2, and through D draw 
DN II OX. With as a center and 
3 as a radius draw an arc cutting 
DN in some point as P. Draw 
PiW_L0X, and draw OP; then 
esc XOP = 3/2 =csc^l. 
.-. ZX0P = ^4. 

OM = V32 - 22 = V5. 
.-. sin A = 2/3, esc A = 3/2 ; 
cos A = V5/3, sec ^4 = 3/V5 ; 
tan A = 2/V5, cot A = V5/2. 
J. = 41° 48' 30". 

13. When tan J. = 4. 
Here MP/OM= 4/1. 
Hence if OM = 1, ifP = 4. 

On OX lay off OM = 1 , and at 
M draw .MS _L OX On MS lay off 
JlfP = 4, and draw OP ; then 
tanXOP = 4 = tan^4. 

.-. Z XOP = JL 

0P = V42 + 12 = V17. 
.-. sin A = 4/V17, esc ^4 = VI 7/4 ; 
cos A = 1/V17, sec A = VI 7 ; 
tan^4 = 4, cot^l = l/4. 

A = 75° 57' 49". 



PLANE TRIGONOMETRY 



14. When cot J. = 7. 
Here OM/MP = 7/1. 
Hence if MP = 1, OM = 7. 
On OX lay off OM - 7, and at 
Jf draw MS ± OX. On Jf S lay off 
IfP = 1, and draw OP ; then 
cot XOP = 7 = cot .A. 

.-. Z XOP = A. 

OP = V72 + 12 = 5 V2. 
.-. sin ^4 = 1/(5 V2), csc ^4 = 5 V2 ; 
cos ^4 = 7/(5 V2) , sec J. = 5 V2/7 ; 
tanJ. = l/7, cot^4=7. 

A = 8° T 48". 



15. When tan A = 9. 
Here MP/OM= 9/1. 
Hence if OJf =1, JlfP = 9. 
On OX lay off Oikf = 1, and at 
ikf draw MS _L OX. On .MS lay off 
MP = 9, and draw OP ; then 
tanX0P = 9 = tanJ.. 
.-. ZXOP = A. 

OP = V92 + 1^ = V82. 

.-. sin A = 9/V82, esc A = V82/9 ; 

cos A = 1/V82, sec A = V82 ; 

tan^L = 9, cot A = 1/9. 

.4 = 83° 39' 35". 



16. Express each of the trigonometric ratios of an acute angle A in 
terms of its sine, writing (sin .A) 2 in the form sin 2 A. 
In fig. 2, let OP = 1. 



Then MP/ OP 

Whence MP = sin A, 
and 



ilfP/1, i.e. sin A is the measure of jYP. 



OM = ^/oP 2 -MP 2 



= vl — sin 2 .4. 
Hence cos ^4 = O M/OP 

= Vl- sin 2 A. 



secA = l/Vl-sin 2 A. § 

tan J. = MP/OM 

= si n^/Vl — sin 2 .A. 
cot ^4 = Vl — sin 2 A /sin A. 
esc A = OP /MP = 1/sin ^4. 




Fig. 2 



17. Express each of the trigonometric ratios of an acute angle in terms 



of its cosine. 

In fig. 2, let 

Then 

Whence 
and 

Hence 



0P = 1. 

OM/OP = 0M/1, i.e 

OM = co s A, 

MP = \ / ~0P' 2 -0M' 2 _: 
sin A = JfP /OP = Vl 
.-. esc J. = 1/Vl - cos 2 A. 



cos A is the measure of OM. 

Vl - c os 2 ^.. 
cos 2 A. 



tan .4 = M P/ O M — Vl - co s 2 J. /cos J.. 
cot J. = cos A/Vl — cos 2 A. 
sec J. = OP/OM = 1/cos ^4. 



EXERCISE II 



18. Express each of the trigonometric ratios of an acute angle in terms 



of its tangent. 

In fig. 2, let 

Then 

Whence 
and 



OM-1. 

MP/OM = MP/1, i.e. tan J. is the measure of MP. 

MP = t&nA, 

OP = Volf 2 + MP 2 = Vl + tan 2 .4. 
.-. sin .A = t an A/Vl + tan 2 A, 
esc A =Vl - f tan 2 A/ta n A, 
cos A = 1 /Vl + tan 2 A, 
sec A = Vl + tan 2 A, 
cot A = 1/tan A. 



EXERCISE II 



1. By § 9, cos 30° equals what? 
sin 60° ? cot 35° ? tan 15° ? sec 85° ? 
esc 76°? sin 73° 14'? cos 65° 43'? 
cos 30° = sin (90° - 30°) = sin 60°. 
sin 60° = cos (90° - 60°) = cos 30°. 
cot 35° = tan (90° - 35°) = tan 55°. 
tan 15° = cot (90° - 15°) = cot 75°. 
sec 85° = esc (90° - 85°) = esc 5°. 
esc 76° = sec (90° - 76°) = sec 14°. 
sin 73° 14' = cos (90° - 73° 14') 

= cos 16° 46'. 
cos 65° 43' = sin (90° - 65° 43') 
= sin 24° 17'. 

2. Given sec A = esc A and 

A < 90°, find the value of A. 
esc A = sec (90° -A). 
.-. sec A = sec (90° -A). 
.: A = 90° -A. 

.-. A = 45°. 

3. Given tan A = cot 2 A, find 
the value of A. 

cot 2^1 = tan (90° -2 A). 
.: tanA = tan (90° -2 A). 
.: A = 90° -2 A. 
.: A = 30°. 



4. Given sin 2 A = cos 3 A, find 
the value of A. 

cos 3 A= sin (90° -3 A). 
.-. sin2A = sin(90°-3A). 
.-. 2A = 90° -3A. 
.: A = 18°. 

5. Given tan (A/2) = cot 2 A, 
find the value of A. 

cot 2 A = tan (90°- 2 A). 
.: tan (A/2) = tan (90° - 2 A). 
.-. A/2 = 90° -2 A. 
.: A = 36°. 

6. Given sec (75° + A) = esc 2 A, 
find the value of A. 

csc2A=sec(90°-2A). 
.-. sec (75° + A) = sec (90° -2 A). 
.: 75° + A = 90° -2 A. 

.-. 3A = 15°, ori = 5°. 

7. Given cot (A +50°) = tan 7 A, 
find the value of A. 

tan7A=cot(90°-7A). 
.-. cot (A + 50°) = cot (90° - 7 A), 
.-. A + 50° = 90° -7 A. 
.-. A = 5°. 



PLANE TRIGONOMETRY 



8. Given sin nA 
the value of A. 



cos mA, find 



cos mA = sin (90° — mA). 
sin nA = sin (90° — mA). 
.: nA = 90° - mA. 
.: A = 90°/(n + m). 



9. Given tan cA - 
find the value of A. 



cot (30° -^4), 



cot (30° - A) = tan (90°- 30° -f A). 
.-. tanc4. = tan(60° + ^4). 
.-. cA = 60° + A. 
.-. A = 60°/(c - 1). 



EXERCISE III 



1. Given C = 90°, A = 25°, 
a = 30 ; to find B, 6, c. 

r £ = 90° - 4. = 05°, 
Formulas i 6 = a cot J., 
lc — a csc^l. 
.-. b = 30 cot 25° 
= 30 x 2.1445 
= 64.335, 
and c = 30 esc 25° 

= 30 x 2.3662 
= 70.986. 

2. Given C = 90°, B = 55°, 
b = 10; to find ^4, a, c. 

A = 90° - J5 = 35°, 
a = 6 cot Z?, 
c = 6 esc B. 
.: a = 10 cot 55° 

= 10 x 0.7002 = 7.002, 
and c = 10 esc 55° 

= 10 x 1.2208 = 12.208. 

3. Given C = 90°, A = 65°, 
c = 70 ; to find J5, a, 6. 

B = 90° - J. = 25°, 
a = c sin ^4, 
6 = c cos J.. 
.-. a = 70 sin 65° 

= 70 x 0.9063 = 63.441, 
and b = 70 cos 65° 

= 70 x 0.4226 = 29.582. 



4. Given C = 90°, B = 15°, 
6 = 20; to find .4, a, c. 

^4 = 90° - B = 75°, 
a = b cot 5, 
c = b esc 2?. 
.-. a = 20 cot 15° 

= 20 x 3.7321 = 74.642, 
and c = 20 esc 15° 

= 20 x 3.8637 = 77.274, 

5. Given C = 90°, B = 35°, 
a = 50 ; to find A, 6, c. 

4 = 90° - B = 55°, 
b = a tan I>, 
c = a sec 5. 
.-. b = 50 tan 35° 

= 50 x 0.7002 = 35.01, 
and c = 50 sec 35° 

= 50 x 1.2208 = 61.04. 

6. Given C = 90°, B = 55°, 
c = 60; to find A, a, b. 

A = 90° - B = 35°, 
a = c cos JB, 
b = c sin _B. 
.-. a = 60 cos 55° 

= 60 x 0.5736 = 34.416, 
and b = 60 sin 55° 

= 60 x 0.8192 = 49.152. 



EXERCISE III 



7. Given C = 90°, a = 36.4, 
6 = 100; to find A, B, c. 

tan A = a/b, 

B = 90° - A, 
c = a csc^4. 
.-. tan .4 = 36.4/100 = 0.3640. 
.-. A = 20°. 
.-. B = 70°, 
and c = 36.4 esc 20° 

= 36.4x2.9238=106.4. 



11. Given C = 90°, B = 75°, 
c = 40; to find A, a, b. 

A = 90°-B = 15°, 
a = c sin A, 
b = c cos ^4. 
.-. a = 40 sin 15° 

= 40 x 0.2588 = 10.352, 
and b = 40 cos 15° 

= 40 x 0.9659 = 38.636. 



8. Given C = 90°, a = 23.315, 


12. Given C = 90°, A = 10 c 


6 = 50 ; to find A, B, c. 


6 = 30 ; to find a, 5, c. 


tan A = a/b, 


B=90° -A = 80°, 


B = 90° - ^4, 






a = & tan A , 


c = bsecA. 




.-. tan J. = 23.315/50 = 0.4663. 


c = bsecA. 


.-. A = 25°. 


.-. a = 30 tan 10° 


.-. B = 65°, 


= 30 x 0.1763 = 5.289, 


and c = 50 sec 25° 


and c = 30 sec 10° 


= 50 x 1.1034 = 55.17. 


= 30 x 1.0154 = 30.462 



9. Given C = 90°, b = 93.97, 
c = 100: to find A, a, B. 

cos A = b/c, 

B = 90°- A, 
a = c sin J.. 
.-. cos^. = 93.97/100 = .9397. 
.-. J. = 20°. 
.-. B = 70°, 
and a = 100 sin 20° 

= 100 x 0.3420 = 34.2. 

10. Given C = 90°, a = 17.1, 
c = 50 ; to find J., B, b. 
sin J. = a/c, 
B = 90° - ^4, 
& = c cos .4. 
.-. sin 4. = 17.1/50 = .3420. 
.-. A = 20°. 
.-. 5 = 70°, 
and b = 50 cos 20° 

= 50 x 0.9397 = 46.985. 



13. Given C = 90°, A = 20°, 
c = 80 ; to find a, B, b. 

B = 90° - A = 70°, 
a = csin^4, 
6 = c cos A. 
.-. a = 80 sin 20° 

= 80 x 0.3420 = 27.36, 
and b = 80 cos 20° 

= 80 x 0.9397 = 75.176. 

14. Given C = 90°, B = 25°. 
a = 30 ; to find A, b, c. 

A = 90°-B = 65°, 
b — a tan B, 
c = a sec 5. 
.-. b = 30 tan 25° 

= 30 x 0.4663 = 13.989, 
and c = 30 sec 25° 

= 30 x 1.1034 = 33.102. 



O PLANE TRIGONOMETRY 

15. Given C = 90°, a = 30.21, 16. Given C = 90°, a = 13.4, 

c = 331 ; to find A, B, b. b = 50 ; to find A, B, c. 

sin A = a/c, tan A = a/b, 

B = 90°-A, B = 90°-A, 

b = ccosA. c — a esc A. 

.-. sin A = 30.21/331 = .9063. .-. tan A = 13.4/50 = .2680, 

..-. A = 65°. .-. A = 15°. 

.-. B = 25°, .-. B = 75°, 

and b = 33± cos 65° = 14.087. and c = 13.4 esc 15° 

= 13.4 x 3.8637 = 51.77. 
EXERCISE IV 

1. The length of a kite string is 250 yd. and the angle of elevation 
of the kite is 40°. Find the height of the kite, supposing the line of the 
kite string to be straight. 

Let A = the angle of elevation, c = the length of the string, and a = the 
height of the kite ; then c = 250 and A = 40°, to find a. 

a- c sin A =250 sin 40° = 250(0.6428) = 160.7. 
Hence the height of the kite is 160.7 yd. 

2. A stick 10 ft. in length stands vertically in a horizontal area, and the 
length of its shadow is 8.391 ft. Find the angle of elevation of the sun. 

Let a = the length of the stick, and b = the length of the shadow ; then 
a = 10 and b = 8.391, to find A. 

ten A = a/b = 10/5.391 = 1.1918. 
.-. A = 50°. 
Hence the elevation of the sun is 50°. 

3. A tree is broken by the wind so that its two parts form with the 
ground a right-angled triangle. The upper part makes an angle of 35° 
with the ground, and the distance on the ground from the trunk to the 
top of the tree is 50 ft. Find the length of the tree. 

Let A = the angle the upper part of the tree makes with the ground, 
and b = the distance from the trunk to the top ; then, in A ABC, A = 35° 
and b = 50, to find a + c. 

a = frtan^l, 
c = bsecA. 
.-. a = 50 tan 35° = 50 (0.7002) = 35. 01, 
and c = 50 sec 35° = 50(1.2208) = 61.04. 

.-. a + c = 96.05. 
Hence the length of the tree is 96.05 ft. 



EXERCISE IV 9 

4. The distance between two towers on a horizontal plane is 60 ft. , and 
the angle of depression of the top of the first as seen from the top of the 
second, which is 150 ft. high, is 25°. Find the height of the first tower. 

Let b = the distance between the towers, and A = the angle of depres- 
sion ; then A = 25° and b = 60, to find 150 — a. 

a = btsn\A = 60 tan 25° 

= 60 (0.4663) = 27.98. 
.-. 150 - a = 122.02. 

Hence the height of the first tower is 122.02 ft. 

5. At a point 200 ft. from the base of an unfinished tower the angle 
of elevation of its top is 20° ; when completed, the angle of elevation of its 
top at this point will be 30°. How much higher is the tower to be built ? 

Let C be the base of the tower, A the point 200 ft. from the base, B 
the top of the unfinished tower, and B' the top of the finished tower ; then 
in A ABC, ZCAB =20°, AC = 200\ . _ . __, _ p 

and in A AB'C, A CAB' = 30°, A C = 200 / t0 find CB ~ CR 
CB = AC tan CAB, 
CB' = ACt<mCAB'. 
.: CB = 200 tan 20° = 200 (0.3640) = 72.80, 
and CB' = 200 tan 30° = 200 (0.5774) = 115.48. 

.-. CB' -CB = 115.48 - 72.80 = 42.68. 

6. The angle of elevation of the sun is 65° and the length of a tree's 
shadow on a level plane is 50 ft. Find the height of the tree. 

Let C be the foot of the tree, B the top, and A the end of the shadow; 
then in A ABC we have A = 65° and b = 50, to find a. 

a = b tan A = 50 tan 65° = 50 (2. 1445) = 107. 22. 
Hence the height of the tree is 107.22 ft. 

7. A chimney stands on a horizontal plane. At one point in this 
plane the angle of elevation of the top of the chimney is 30° ; at another 
point 100 ft. nearer the base of the chimney the angle of elevation of 
the top is 45°. Find the height of the chimney. 

Let C be the base of the chimney, B the top, A the first point, and 
A' the second point of observation ; then A A' = 100 ft., Z CA'B = 45°, 
and Z CAB = 30°, to find CB. 

AC/CB = cot 30° = 1.7321. 
A'C/CB = cot 45° = 1.0000. 

.-. 100 = AC - A'C = (1.7321 - 1) CB. 
.-. CB = 100/. 7321 = 136.6. 
Hence the height of the chimney is 136.6 ft. 



10 PLANE TRIGONOMETRY 

8. A person standing on the bank of a river observes that the angle 
subtended by a tree on the opposite bank is 50°; walking 40 ft. from the 
bank he finds the angle to be 30°. Find the height of the tree and the 
breadth of the river, if the two points of observation are in the same 
horizontal line at the base of the tree. 

Let C be the foot and B the top of the tree, A the first and A' the second 
point of observation ; then A' A = 40, Z CAB = 50°, and Z CA'B = 30°, 
to find CB and AC. 

A'C/CB = cot 30° = 1.7321. 
AC/CB = cot 50° = 0.8391. 

.-. 40 = A'C -AC = (1.7321 - .8391) CB. 
.: C£ = 40/.8930 = 44.79. 
Also AC=CB cot 50° = 44. 79 (0. 8391) = 37. 58. 

Hence the height of the tree is 44.79 ft. and the breadth of the river is 
37.58 ft. 

9. The shadow of a tower standing on a horizontal plane is found to 
be 60 ft. longer when the sun's altitude is 30° than when it is 45°. Find 
the height of the tower. 

Let C be the base and B the top of the tower, A the end of the shadow 
when the sun's altitude is 45° and A' the end when the altitude is 30° ; 
then A' A = 60 ft., Z CAB = 45°, and Z CA'B = 30°, to find CB. 

A'C/CB = cot 30° = 1.7321. 
AC/CB = cot 4o° = l. 

.-. 60 = A'C -AC = (1.7321 - 1) CB. 
.-. CB = 60/.7321 = 81.96. 

Hence the height of the tower is 81.96 ft. 

10. At a point midway between two towers on a horizontal plane the 
angles of elevation of their tops are 30° and 60° respectively. Show that 
one tower is three times as high as the other. 

Let C and B' be the base and top of the higher tower, C and B the 
base and top of the other, and A the point midway between the towers ; 
then Z CAB = 30° and Z CAB' = 60°, to prove that C'B'/CB = 3. 

C'B'/A C = C'B'/AC = tan 60° = V3. 
CB/AC = tan 30° = 1/V3 = V3/3. 
.-. CB' :CB = ^S: V3/3 = 3:1. 



EXERCISE IV 11 

11. Two observers on the same horizontal line and in the same vertical 
plane with a balloon, on opposite sides of it and 2500 ft. apart, find its 
angles of elevation to be 35° and 55° respectively. Find the height of the 
balloon. 

Let A and A' be the points of observation, B the position of the balloon, 
and C the foot of the vertical line from B ; then J.J/=2500, /. CAB=3b°, 
and ZCA'B = 55°, to find CB. 

AC/CB = cot 35° = 1.4281. 
CA'/CB = cot 55° = .7002. 

.-. 2500 = AC + CA' = (1.4281 + .7002) CB. 
.-. CB = 2500/2.1283 = 1174.6. 

Hence the height of the balloon is 1174.6 ft. 

12. A man in a balloon observes that the bases of two towers, which 
are a mile apart on a horizontal plane, subtend an angle of 70°. If he is 
exactly above the middle point between the towers, find the height of the 
balloon. 

Let A and A' be the bases of the two towers, B the balloon, and C the 
foot of the vertical line from B; then AC = \ mile and Z. ABC = 35°, to 

find CR CB/AC = cotS5°. 

.-. CB = (| cot 35°) mi. = .7141 mi. = 3770 ft. 

13. From the foot of a tower the elevation of the top of a church 
spire is 55°, and from the top of the tower, which is 50 ft. high, the 
elevation is 35°. Find the height of the spire and the distance of the 
church from the tower, if both stand on the same horizontal plane. 

Let A be the foot of the tower and B its top, and C the foot of the 
church spire and D its top; then CD and AC are required. 

tan 55° = CD/ AC. 

cot 35° = AC /(CD - 50). 

.-. tan 55° cot 35° = CD/(CD - 50). 

nTk 50 tan 55° cot 35° 50 x 1.4281 x 1.4281 _ Q Ano 

.'. CD = = = 98.098. 

tan 55° cot 35° - 1 (1.4281) 2 - 1 

Also AC = CD cot 55° = 98.098 x .7002 = 68.69. 

14. From the top of a tower whose height is 108 ft. the angles of 
depression of the top and bottom of a vertical column standing on a level 
with the base of the tower are found to be 25° and 35° respectively. 
Find the height of the column and its distance from the tower. 



12 PLANE TRIGONOMETRY 

Let A be the foot and B the top of the tower, F the foot and T the top 
of the column ; then FT and AF are required. 

tan 35° = 108/ AF. 
cot 25° = AF/ (10$ - FT). 
.-. tan 35° cot 25° = 108/(108 - FT). 
_ 108 (tan 35° cot 25° - 1) _ 108 (.7002 x 2.1445 - 1) _ 
tan 35° cot 25° ~ . 7002 x 2. 1445 

Also AF = 108 cot 35° = 108 x 1.4281 = 154.23. 

15. Two pillars of equal height stand on opposite sides of a horizontal 
roadway which is 100 ft. wide. At a point in the roadway between the 
pillars the angles of elevation of their tops are 50° and 25° respectively. 
Find the height of the pillars and the position of the point of observation. 

Let C be the point in the roadway, and A the foot and B the top of 
the pillar farthest away; then AB and AC are required. 

%2^2b° = AB/CA. 
cot 50° = (100 - CA)/AB. 
.-. tan 25° cot 50° = (100 - CA)/CA. 

,. CA = ™ - = ™ = 71.88. 

tan 25° cot 50° + 1 . 4663 x . 8391 + 1 

Also AB = CA tan 25° = 71.88 x .4663 = 33.52. 

16. A house 50 ft. high and a tower stand on the same horizontal 
plane. The angle of elevation of the top of the tower at the top of the 
house is 25°, on the ground it is 55°, Find the height of the tower and 
its distance from the house. 

Let A be the point on the ground by the house and B the top of the 
house, F the base of the tower and T its top; then FT and AF are 
required. tan 25 o _ (Fr _ 50 y AK 

cot 55° = AF/FT. 

.: tan 25° cot 55° = {FT - 50)/FT. 

, FT = *> = 52 = 7 4.24. 

(1- tan 25° cot 55°) 1 - .4663 x .7002 

Also AF= FT cot 55° = 74.24 x .7002 = 51.98. 

17. On the top of a bluff is a tower 75 ft. high ; from a boat on the 
bay the angles of elevation of the top and base of the tower are observed 
to be 25° and 15° respectively. Find the horizontal distance of the boat 
from the tower, also the distance of the boat from the top of the tower. 



EXERCISE IV 



13 



Let A be the position of the boat, T the top and B the base of the 
tower, and C the intersection of the vertical line through B with the 
horizontal plane through A ; then A C and A T are required. 
CB=ACtSLnlb°. 
.-. CT = AC tan 15° + 75. 
AC= CT/tan 25° = (AC tan 15° + 75) /tan 25°. 
.-. ^C = 75/(tan 25° - tan 15°) 

= 75/(.4663 - .2679) = 378.02. 
Also A T= AC sec 25° = 378.02 x 1.1034 =417.11. 

18. One of the equal sides of an isosceles triangle is 50 ft. and one of 
its equal angles is 40°. Find the base, the 
altitude, and the area of the triangle. 

In this and the next five examples we 
shall use the following notation : 

r = one of the equal sides, 
c = base, 
h = altitude, 

A = one of the equal angles, 
C = angle at the vertex, 
Q = area of the triangle. 

Here r = 50 and A = 40°, to find c, h, Fig. 3 

and Q. h = r sin A = 50 sin 40° = 50 x .6428 = 32.14. 

c = 2 r cos ^4 = 100 cos 40° = 100 x .7660 = 76.60. 
Q = ch/2 = 76.6 x 32.14/2 = 1231. 

19. The base of an isosceles triangle is 68.4 ft. and each of its equal 
sides is 100 ft. Find the angles, the height, and the area. 

Here c = 68.4, r = 100, to find A, C, h, Q. 
cos A = (c/2)/r = 34.2/100 = .342. 
.-. A = 70°. 

C = 2 (90° -A) = 40°. 

h = r sin A = 100 x .9397 = 93.97. 

Q = ch/2 = 68.4 x 93.97/2 = 3213.8. 

20. The base of an isosceles triangle is 100 ft. and its height is 35.01 ft. 
Find its equal sides and the angles. 

Here c = 100, h = 35.01, to find r, A, C. 

tan A = 2 h/c = 2 x 35.01/100 = .7002. 
.-. A = 35°. 

C = 2(90° -A) = 110°. 

r = hcscA = 35.01 esc 35° = 35.01 x 1.7434 = 61.04. 




14 PLANE TRIGONOMETRY 

21. The base of an isosceles triangle is 88 ft. and its vertical angle is 
70°. Find the height, the equal sides, and the area. 

Here c = 88, C = 70°, to find h, r, Q. 

h = (c/2)cot(C/2) = 44 cot 35° = 44 x 1.4281 = 62.836. 
r = (c/2) csc (C/2) = 44 esc 35° = 44 x 1.7434 = 76.71. 
Q = ch/2 = 88 x 62.836/2 = 2764.8. 

22. The base of an isosceles triangle is 100 ft. and the equal angles 
are each 65°. Find the equal sides, the height, and the area. 

Here c = 100, A = 65°, to find r, h, Q. 

h = (c/2) tan A = 50 tan 65° = 50 x 2.1445 = 107.23. 
r = (c/2) sec A = 50 sec 65° = 50 x 2.3662 = 118.3. 
Q = ch/2 = 100 x 107.23/2 = 5361.5. 

23. The height of an isosceles triangle is 50 ft. and its vertical angle 
is 60°. Eind the sides and the area. 

Here h = 50, C = 60°, to find r, c, Q. 

r = hsec (C/2) = 50 sec 30° = 50 x 1.1547 = 57.74. 
c ,= 2 h tan (C/2) = 100 tan 30° = 100 x .5774 = 57.74. 
Q = ch/2 = 57 .74 x 50/2 = 1443.5. 

24. A man's eye is on a level with and 100 ft. distant from the foot 
of a flag pole 36.4 ft. high. When he is looking at the top of the pole, 
what angle doe # s his line of sight make with a line from his eye to the 
foot of the pole ? 

Let C be the foot of the flag pole, B its top, and A the point of obser- 
vation ; then Z CAB is required. 

tan CAB = CB/AC = 36.4/100 = 0.3640. 
.-. Z CAB = 20°. 

25. A circular pond has a pole standing vertically at its center and its 
top is 100 ft. above the surface. At a point in the circumference the 
angle subtended by the pole is 20°. Find the radius and the area of the 
pond. 

Let C be the foot of the pole, B its top, and A the point in the circum- 
ference ; then A C and 7tA C 2 are required. 

AC T = BC cot CAB = 100 cot 20° = 274. 75. 
it AC 2 = 3.1416(274.75) 2 = 237150. 

Note. In the value of 7t AC 2 , the significant figures after the fifth are 
unreliable, and are therefore not retained. 



EXERCISE IV 15 

26. A ladder 33| ft. long leans against a house and reaches to a point 
30.21 ft. from the ground. Find the angle between the ladder and the 
house and the distance the foot of the ladder is from the house. 

Let A be the bottom of the ladder, B its top, and C the intersection of 
the vertical line from B with the ground; then A CBA and AC are 
required. 

cos CBA = CB/AB = 30.21/33*- = .9063. 
.-. Z CBA = 25°. 

AC =CB tan CBA = 30.21 tan 25° = 30.21 x .4663 = 14.09. 

27. From the summit of a hill there are observed two consecutive 
milestones on a straight horizontal road running from the base of the 
hill. The angles of depression are found to be 10° and 5° respectively. 
Find the height of the hill. 

Let A and A / be the two consecutive milestones, B the top of the hill, 
and C the foot of the vertical line from B to the horizontal plane through 
A and A" ; then CB is required. 
A'C = CB cot 5°. 
AC = CB cot 10°. 

5280 = A'C -AC=CB (cot 5° - cot 10°). 
.-. CB = 5280/(cot 5° - cot 10°) = 5280/(11.4301 - 5.6713) = 916.86. 

28. At the foot of a hill the angle of elevation of its summit is observed 
to be 30° ; after ascending the hill 500 ft., up a slope of 20° inclination, 
the angle of elevation of its summit is found to be 40°. Find the height 
of the hill if the two points of observation and the summit are in the same 
vertical plane. 

Let A be the first and A' the second point of observation, B the top of 
the hill, and C and C the points in which the vertical line from B inter- 
sects the horizontal planes through A and A' respectively ; then CB is 
required. 

Draw A' H ± AB ; then 

A'H/oOO = sin 10°. 
A'B/A'H = esc A'BH = esc 10°. 
C'B/A'B = sin 40° = .6428. 

Multiplying together these three equations, we obtain 
C'B/oOO = .6428 esc 10° sin 10° = .6428. 
.-. C'B = 321.4. 

CC = 500 sin 20° = 500 x .3420 = 171. 
.-. CB = CC' + CB = 321.4 + 171 = 492.4. 



16 PLANE TRIGONOMETRY 

29. At the foot of a mountain the angle of elevation of its summit is 
35° ; after ascending an opposite mountain 3000 ft. , up a slope of 15° 
inclination, the angle of elevation of the summit is 15°. Find the height 
of the first mountain if the points of observation and the summit are in 
the same vertical plane. 

Let A be the first and A' the second point of observation, B the top of 
the hill, and C and C the points in which the vertical line from B inter- 
sects the horizontal planes through A and A' respectively ; then CB is 
required. 

Draw A'H J_ A B ; then Z HA A' = Z CAB + 15° = 50°. 
A'H/ 3000 = sin HA A' = sin 50° = .7660. 
A'B/A'K = esc A'BH = esc 20° = 2.9238. 
C'B/A'B = sin 15° = .2588. 

.-. CB = 3000 x .766 x 2.9238 x .2588 = 1738.8. 

CC = 3000 sin 15° = 3000 x .2588 = 776.4. 
.-. CB ^ CC + CB = 2515. 

30. From the extremities of a ship 500 ft. long the angles which the 
direction of a buoy makes with that of the ship are 60° and 75°. Find the 
distance of the buoy from the ship, having given that cot 75° = 2 — V3. 

Let A and B be the ends of the ship and D the buoy. Draw CD ± AB ; 
then CI) is the distance required. 
^iC/CD = cot60 o . 
C5/CD = cot 75°. 

... ^^ = ^ = cot60o + cot75o = l V 3 + 2-V3. 

,CB= 5Q0 /Q = ^g- =12 5 ( a + V 3). 

2-| V3 3-V3 V v . 

31. There are two posts which are 240 and 80 ft. high respectively. 
From the foot of the second the elevation of the top of the first is found 
to be 60°. Find the elevation of the second from the foot of the first. 

Let AB be the first post and CD the second ; then Z CAD is required. 
AC = AB cot 60° = 240 x 0. 5774 = 138. 57. 
cot CAD = AC/CD = 138.57/80 = 1.7321. 
.-. Z CAD = 30°. 

32. A boy standing c feet behind and opposite the middle of a football 
goal sees that the angle of elevation of the nearer crossbar is J., and the 
angle of elevation of the farther one is B. Show that the length of the 
field is c (tan A cot B — 1). 



EXERCISE IV 17 

Let H be the position of the boy, HE and HG the lines of sight to the 
crossbars, and B and F the feet of the vertical lines from E and G ; then 
DF is required. 

FG = DE = ctan A. 

HF = FG cot B = c tan A cot B. 
.-. BF= HF -HB = c (tan J. cot 5 - 1). 

33. A valley is crossed by a horizontal bridge whose length is I. The 
sides of the valley make angles A and B with the horizon. Show that 
the height of the bridge above the bottom of the valley is l/(cot A + cot B). 

Let ST be the bridge and R the lowest point of the valley under the 
bridge. Draw RD ± ST; then RD is required. 

SB/BR = cot A. 
I)T/DR = cotB. 
.-. (SB + BT)/BR = l/BR = cot A + cot 5. 
.-. BR = l/(cotA + cot B). 

34. Two forces of a and 6 lbs. respectively act in the same direction. 
Find their resultant. Illustrate the problem geometrically. 

The resultant is a force of a + b lbs. acting in the same direction as the 
two forces. 

The two forces can be represented geometrically by two lines a units 
and b units long respectively and having the direction of the forces. 
The sum of these lines will be a line a + b units long and having the 
direction of the first lines. 

35. Two forces of a and b lbs. respectively act in opposite directions. 
Find their resultant when a > b, when a = 6, and when a < b. Illustrate 
each case geometrically. 

When a > b the resultant is a force of a — b lbs. acting in the same 
direction as a. When a — b the resultant is 0. When a < b the result- 
ant is a force of b — a lbs. acting in the same direction as 6. 

The first force can be represented geometrically by a line a units long, 
and having the direction of the force ; and the second by a line b units 
long, and having the direction of the second force. When a > b the sum 
of these two lines will be a line a — b units long, and having the direction 
of the first line. When a = b the sum of the lines will be 0. When a < b 
the sum of the lines will be a line b — a units long, and having the direction 
of the second line. 



18 PLANE TRIGONOMETRY 

' 36. By two or more experiments verify that if in any triangle ABC 
the two sides AB and BC represent two forces (both in size and direction), 
the third side AC will represent their resultant, i.e. their sum in its 
simplest form. 

Tie three strings together ; to each loose end attach a spring balance ; 
fasten the other ends of these three balances to three screw hooks, two 
in the upper frame and one on the lower frame of the blackboard, bring- 
ing any desired tension on the balances. Project on the board the three 
strings and lay off on each line a length representing the tension of the 
string projected, as read from the spring balance. Place the origin of one 
of these directed lines at the end of a second of these lines ; then the 
line whose origin is that of the first line and whose end is that of the 
second line will be the negative of the third line. Hence the directed lines 
representing the tensions on the three strings are the sides of a triangle. 

This is one of the simplest methods of making this experiment. 

37. Two forces of 3 and 4 lbs. respectively act at right angles to each 
other. Show that their resultant is a force of 5 lbs. and that its line of 
action and that of the first force make an angle whose tangent is 4/3. 

Let AC represent the first force and CB the second; then AB will 
represent their resultant. 

Now the length of AB = V3 2 + 4 2 = 5, 
and Z CAB = the angle whose tangent is 4/3. 

38. Two forces of a and b lbs. respecti vely act at right angles. Show 
that their resultant is a force of V a 2 + b 2 lbs. , and that its line of action 
and that of the first force make an angle whose tangent is b/a. 

Let AC represent the first force and CB the second ; then AB will 
represent their resultant. 

Now the length of AB = Va 2 + &i 
and Z CAB = the angle whose tangent is b/a. 

39. Two forces act at right angles. The first is a force of 3 lbs. and 
the resultant is one of 5 lbs. Show that the second force is one of 4 lbs., 
and that the lines of action of the first force and the resultant form an 
angle whose cosine is 3/5. 

Let A C represent the first force and AB the resultant ; then CB will 
represent the second force. 

Now the length of CB = V5 2 - 3 2 = 4, 
and Z CAB = the angle whose cosine is 3/5. 



EXERCISE V 19 

40. Two forces act at right angles. The first is a force of a l bs, and the 
resultant is one of c lbs. Show that the second force is one of Vc 2 — a 2 lbs. 
and that the lines of action of the first force and the resultant form an 
angle whose cosine is a/c. 

Let AC represent the first force and AB the resultant; then CB will 
represent the second force. 

Now the length of CB = Vc^o 2 , 
and Z CAB = the angle whose cosine is a/c. 

41. Construct an angle of 6° or 84° and prove that 

sin 6° =cos84°>0.1, 

tan 6° =cot84°>0.1, 
and tan 84° = cot 6° < 10. 

By § 9, sin 6° = sin (90° - 84°) = cos 84°, 

tan 6° = tan (90° - 84°) = cot 84°, 
and tan 84° = tan (90° - 6°) = cot 6°. 

Draw Z XOB = 6° ; on OB take OP = 10, and draw PM _L OX. 
By measurement MP > 1 ; hence sin 6° = MP /OP > 0. 1. 

Since OM < OP, tan 6° = MP/OM > MP/ OP > 0. 1, 
and cot 6° = OM/MP < OP /MP < 10. 

EXERCISE V 

1. In which quadrant is an angle of 5/3 right angles ? 
In the second quadrant. 

2. In which quadrant is an angle of 3f right angles ? 
In the fourth quadrant. 

3. In which quadrant is an angle of 17| right angles ? 
In the second quadrant. 

4. In which quadrant is an angle of 150° ? 317° ? 

An angle of 150° is in the second quadrant, and one of 317° is in the 
fourth quadrant. 

5. In which quadrant is an angle of 847° ? 1111° ? 

,847° = 2 x 360° + 127° ; hence an angle of 847° is coterminal with one 
of 127°, and is therefore in the second quadrant. 1111° = 3 x 360° + 31° ; 
hence an angle of 1111° is coterminal with one of 31°, and is therefore in 
the first quadrant. 



20 PLANE TRIGONOMETRY 

6. In which quadrant is an angle of — 35° ? — 140° ? 

An angle of —35° is in the fourth quadrant, and one of — 140° is in 
the third quadrant. 

7. In which quadrant is an angle of - 225° ? - 300° ? 

An angle of — 225° is in the second quadrant, and one of — 300° is in 
the first quadrant. 

8. In which quadrant is an angle of — 415° ? — 842° ? 

- 415° = (- 1)360° + (- 55°); hence an angle of - 415° is coterminal 
with one of — 55°, and is therefore in the fourth quadrant. 

- 842° = (- 2)360° + (- 122) ; hence an angle of - 842° is coterminal 
with one of — 122°, and is therefore in the third quadrant. 

9. In which quadrant is an angle of 942° ? - 1174° ? 

942° = 2 x 360° + 222° ; hence an angle of 942° is coterminal with one 
of 222°, and is therefore in the third quadrant. 

-1174°= (-3)360°+ (-94°); hence an angle of -1174° is coterminal 
with one of — 94°, and is therefore in the third quadrant. 

10. Construct an angle of 847° ; 1111°; -225°; -300°; 942°; -1174°. 
847° = 2 x 360° + 127° ; hence construct an angle of 127°, and one of 

847° will be coterminal with it. 

1111° = 3 x 360° + 31°; hence construct an angle of 31°, and one of 
1111° will be coterminal with it. 

An angle of — 225° is coterminal with one of 135° ; hence to construct 
an angle of — 225° construct one of 135°. 

An angle of — 300° is coterminal with one of 60° ; hence to construct 
an angle of — 300° we construct one of 60°. 

942° = 2 x 360° + 222° ; hence to construct an angle of 942° construct 
one of 222° or one of - 138°. 

- 1174° = (- 3)360° + (- 94°); hence to construct an angle of -1174° 
construct one of — 94°. 

11. Give two positive and two negative angles, each of which is coter- 
minal with 45° ; 30° ; 100° ; 200° ; - 10° ; - 100°. 

An angle of 405°, 765°, - 315°, or - 675° is coterminal with one of 45°. 
An angle of 390°, 750°, - 330°, or - 690° is coterminal with one of 30°. 
An angle of 460°, 820°, - 260°, or - 620° is coterminal with one of 100°. 
An angle of 560°, 920°, - 160°, or - 520° is coterminal with one of 200°. 
An angle of 350°, 710°, - 370°, or - 730° is coterminal with one of -10°. 
An angle of 260°, 620°, - 460°, or - 820° is coterminal with one of - 100°. 



EXERCISE V 21 

12. Find the complement and supplement of 165°. 
The complement of 165° = 90° - 165° = - 75°. 
The supplement of 165° = 180° - 165° = 15°. 

13. Find the complement and supplement of 228°. 
The complement of 228° = 90° - 228° = - 138°. 
The supplement of 228° = 180° - 228° = - 48°. 

14. Find the complement and supplement of 295° 17' 14". 

The complement of 295° 17' 14" = 90° - 295° IT 14" = - 205° 17' 14". 
The supplement of 295° 17' 14" = 180° - 295° 17' 14" = - 115° 17' 14". 

15. Find the complement and supplement of 314° 22' 17". 

The complement of 314° 22' 17" = 90° - 314° 22' 17" = - 224° 22' 17". 
The supplement of 314° 22' 17" = 180° - 314° 22' 17" = - 134° 22' 17". 

16. Find the complement and supplement of — 32° 14' 21". 

The complement of - 32° 14' 21" = 90° + 32° 14' 21" = 122° 14' 21". 
The supplement of - 32° 14' 21" = 180° + 32° 14' 21" = 212° 14' 21". 

17. Find the complement and supplement of — 165° 28' 42". 

The complement of - 165° 28' 42" = 90° + 165° 28' 42" = 255° 28' 42". 
The supplement of - 105° 28' 42" = 180° + 165° 28' 42" = 345° 28' 42". 

18. Find the smallest positive angle coterminal with 420°. 

420° = 360° + 60° ; hence an angle of 420° is coterminal with one of 60°. 

19. Find the smallest positive angle coterminal with 895°. 

895° = 2 x 360° + 175°; hence an angle of 895° is coterminal with one 
of 175°.' 

20. Find the smallest positive angle coterminal with — 330°. 
An angle of — 330° is coterminal with one of 30°. 

21. Find the smallest positive angle coterminal with — 740°. 

- 740° = (- 2)360° + (- 20°); hence an angle of - 740° is coterminal 
with one of - 20° or one of 340°. 

22. Find the smallest positive angle coterminal with one of — 1123°. 

- 1123° = (- 3) 360° + (- 43°) ; hence an angle of -1123° is coterminal 
with one of — 43° or one of 317°. 



22 



PLANE TRIGONOMETRY 



EXERCISE VI 



sin -1 (— 2/3), construct A and find its other trigono- 




1. Given A 
metric ratios. 

Here sin A = - 2/3. 

Since sin A is — , A is in the third or 
fourth quadrant. 

Here MP/ OP = - 2/3. 

Hence if OP = 3, MP = - 2. 

Draw FF' ± X'X at 0, as in fig. 4. 

On OF' lay off OD = - 2, and through 
D draw LK II X'X. 

With as a center and 3 as a radius 
draw an arc cutting LK in two points as P and P'. Draw OP and OP' ; 
then A is any angle coterminal with Z XOP or Z. XOP'. 

For draw PM ± X'X and P'Jf' ± X'X ; then sin XOP = - 2/3, and 
sin XOP' = - 2/3. 

0M=- V32 - 22 = - V5, OM' = V5- 

.-. sin A = - 2/3, esc A = - 3/2 ; 

cos vl = T Vo/3, sec J. = t 3/v/5 ; 

tan J. = ± 2/V5, cot 4 = ± vV2. 

2. Given J. = tan -1 (5/2), construct A and find its other trigonometric 
ratios. 

Here tan A = 5/2. 

Since tan A is + , A is in the first or third quadrant. 

Here .MP/ Oilf= 5/2. 

Hence if OM = 2, MP = 5 ; or if OM = - 2, itfP = - 5. 

Draw X'OX, as in fig. 4. 

On OX take Oilf = 2, at if draw JfS ± XX, on .MS lay off MP = 5, 
and draw OP. 

On OX' take OM' = - 2, at Jf draw WS' ± X'X, on ilf'S' lay off 
M'P' = - 5, and draw OP'. 

Then -4 is any angle coterminal with A XOP or ZXOP'. 

For tan XOP = 5/2 and tan XOP' = - 5/(- 2) = 5/2. 

OP = V22 + 5 2 = V29 = OP'. 

.: sin J. = ± 5/V29, esc A = ± V29/5 ; 

cos A = ± 2/V29, sec A = ± V29/2 ; 

tanA = 5/2, cot J. = 2/5. 



EXERCISE VI 23 

3. Given A = tan- 1 (— 3), construct A and find its other trigonometric 
ratios. 

Here tan A = — 3. 

Since tan A is — , A is in the second or fourth quadrant. 
Here MP/OM = - 3 = - 3/1. 

Hence if OM = - 1, MP = 3 ; or if OM = 1, i¥P = - 3. 
Draw X'OX, as in fig. 4. 

On OX' take OJf = - 1, at M draw MS J_ XX, on JfS take MP = 3, 
and draw OP. 

On OX take OJf' = 1 , at M draw MS' i. X'X, on Jf'S' take Jf'P' = - 3, 
and draw OP' '. 

Then A is any angle coterminal with Z XOP or Z XOP' . 
Eor tan XOP = 3/(- 1) = - 3, and tan XOP' = - 3. 
OP = Vl2 + (_3) 2 = V10 = OP'. 
.-. sin A = ± 3/V10, esc A = ± VI 0/3 ; 

cos J. = T 1/V10, sec J. = =fV1 ; 

tan ^4 = - 3, cot A = - 1/3. 

4. Given .A = cos -1 (2/3), construct ^1 and find its other trigonometric 
ratios. 

Here cos A = 2/3. 

Since cos A is + , A is in the first or fourth quadrant. 
Here OM/OP = 2/3. 
Hence if OP = 3, OM = 2. 
Draw X'OX, as in fig. 4. 

On OX take OM = 2, and through M draw SMS' _L X'X. With as 
a center and 3 as a radius draw an arc cutting SMS' in two points as P 
and P'. Draw OP and OP' ; then ^4 is any angle coterminal with Z XOP 
or Z XOP'. 

For cos XOP = 2/3, and cos XOP' = 2/3. 

MP = V32 - 22 = V5, JfP r = -V5. 
.-. sin J. = ±V5/3» esc ^4 = ±3/V5 ; 

cos^. = 2/3, secJ. = 3/2; 

tan J. = ±V5/2, cotA = ± 2/V5. 

5. Given ^4 = siir- 1 (— 7/8), construct ^4 and find its other trigono- 
metric ratios. 

Here sin A = — 7/8. 

Since sin A is — , A is in the third or fourth quadrant. 

Here MP J OP = - 7/8. 



24 PLANE TRIGONOMETRY 

Hence if OP = 8, MP = - 7. 

Draw YY' J_ X'X at 0, as in fig. 4. 

On OY' take OD = - 7, and through D draw LK II X'X. With as a 
center and 8 as a radius draw an arc cutting LK in two points as P and 
P'. Draw OP and OP' ; then A is any angle coterminal with Z XOP or 
Z XOP'. 

For draw P_M"J_ X'X and P'lf ± X'X ; then sin XOP = -7/8, and 
sin XOP 7 = - 7/8. 



0M=- V8 2 - 7 2 = - V15, OJf' = VI 5. 

.-. sin A = - 7/8, cscS = - 8/7 ; 

cos A = T VI 5/8, sec^l=T 8/V15; 

tan -4 = ± 7/V15, cotA = ±^/lo/7. 

6. Given J. = tan -1 7, construct A and find its other trigonometric 
ratios. 

Here tan A =7. 

Since tan S is + , J. is in the first or third quadrant. 

Here MP/OM = 7=7/1. 

Hence if OM = 1, ItfP .= 7; or if OJf = - 1, MP = - 7. 

Draw X'OX, as in fig. 4. 

On OX take Oikf = 1, at M draw lf£ ± X'X, on ATS take MP = 7, and 
draw OP. 

On OX' take OJf = - 1, at W draw ilf'S' JL X'X, on M'S' take 
M'P' = - 7, and draw OP'. 

Then ^L is any angle coterminal with Z XOP or Z XOP'. 

For tan XOP = 7 and tan XOP' = 7. 



OP = VP + 72 = V50 = 5 V2 = OP'. 

.-. sin A = ± 7/(5 V2), csc.4 = ± (5V2)/7; 

cos J. = ±1/(5 V2), sec^ = ±5V2; 

tan A = 7, cot J. = 1/7. 

7. Given J. = cos- 1 (— 3/7), constructs and find its other trigonometric 
ratios. 

Here cos A =- 3/7. 

Since cos J. is — , A is in the second or third quadrant. 

Here OM/OP = - 3/7. 

Hence if OP = 7, QM = - 3. 

Draw X'OX, as in fig. 4. 

On OX' take OM = - 3, and through if draw SS' ± X'X, 



EXERCISE VI 25 

With as a center and 7 as a radius draw an arc cutting SS' in two 

points as P and P'. Draw OP and OP' ; then A is any angle coterminal 

with Z XOP or Z XOP' . For cos XOP = - 3/7, and cos XOP' == - 3/7. 

jtfP = V7 2 - 3 2 = V40, JfP 7 = - V^O. 

.-. sin A = ± VW7, esc J. = ± 7/ V40 ; 

cos A = — 3/7, sec A = — 7/3 ; 

tan J. = T V40/3, cot J. = T 3/ V40. 

8. Given A = cot -1 (5/3), construct A and find its other trigonometric 
ratios. 

Here cot A = 5/3. 

Since cot^i is +, A is in the first or third quadrant. 
Here OM/MP = 5/3. 

Hence if MP = 3, OM = 5 ; or if MP = - 3, OM = - 5. 
Draw X'OX, as in fig. 4. 

On OX take OM = 5, at M draw MS ± OX, on MS take MP = 3, and 
draw OP. 

On OX 7 take OM' = - 5, at if' draw JITS' ± OX', on M'S' take 
3TP' = - 3, and draw OP'. 
Then A is any angle coterminal with Z XOP or Z XOP'. 
For cot XOP = 5/3, and cot XOP' = 5/3. 

OP = V5 2 + 3 2 = V34 = OP'. 
.-. sin J. = ± 3/V34, csc^L = ± V34/3 ; 

cos A = ± 5/V34, sec A - ± V34/5, 

tan^l = 3/5, cot A = 5/3. 

9. Given A = cos _1 ( — 4/5), construct A and find its other trigono- 
metric ratios. 

Here cos A — — 4/5. 

Since cos A is — , A is in the second or third quadrant. 
Here OM/OP = - 4/5. 
Hence if OP = 5, OJf = - 4. 
Draw X'OX, as in fig. 4. 

On OX' take OM = - 4, and through M draw SS' _L X'X. 
With as a center and 5 as a radius draw an arc cutting SMS' in two 
points as P and P'. Draw OP and OP' ; then A is any angle coterminal 
with Z XOP or Z XOP'. For cos XOP = - 4/5, and cos XOP' = - 4/5. 
MP = V5 2 - 4 2 = 3, MP' = - 3. 
.-. sin J. = ± 3/5, csc^i = ± 5/3; 

cos A = — 4/5, sec A = — 5/4 ; 

tan^L=T3/4, cot^4=T4/3. 



26 PLANE TRIGONOMETRY 

10. Given A = sec _1 2, construct A and find its other trigonometric 
ratios. 

Here sec A = 2. 

Since sec A is + , A is in the first or fourth quadrant. 

Here 0P/0M=2 = 2/1. 

Hence if OP =2, 0M=1. 

Draw X'OX, as in fig. 4. 

On OX take OM = 1, and through Jf draw S3' J_ X'X. 

With as a center and 2 as a radius draw an arc cutting SS' in two 
points as P and P'. Draw OP and OP' ; then A is any angle coterminal 
with Z XOP or Z XOP'. For sec XOP = sec XOP' = 2. 



ifP = V22 - 12 = V3, 2fP' = - V3. 

.-. sin A = ± V3/2, esc A = ± 2/V3 ; 

cos A = 1/2, sec^i = 2; 

tan J. = ± V3, cot A = ± 1/V3. 

11. Given A = sec _1 (- 3/2), construct A and find its other trigono- 
metric ratios. 

Here sec A = - 3/2. 

Since sec A is — , A is in the second or third quadrant. 

HereOP/OM=3/(-2). 

Hence if OP = 3, OM = - 2. 

Draw X'OX, as in fig. 4. 

On OX' take OJf = - 2, and through M draw SS' _L X'X. 

With as a center and 3 as a radius draw an arc cutting SS' in two 
points as P and P'. Draw OP and OP' ; then A is any angle coterminal 
with Z XOP or Z XOP'. For sec XOP = sec XOP' = - 3/2. 

MP = V3 2 - 2 2 = V5, Jf P' = - V5. 

.*. sin J. = ± V5/3, esc ^4 = ± 3/ V5 5 

cos J. = -2/3, sec ^4 = -3/2; 

tan J. = =F V5/2, cot J. = ^ 2/V5. 

12. Given J. = csc- 1 (— 5/3), construct A and find its other trigono- 
metric ratios. 

Here esc A = — 5/3. 

Since esc ^4 is — , A is in the third or fourth quadrant. 

Here OP /MP = 5/ (-3), 

Hence if OP = 5, MP = - 3. 

Draw OF' _L X'X at 0, as in fig. 4. 

On OY' take OD = - 3, and through D draw liT || X'X. 



EXERCISE VII 



27 



With as a center and 5 as a radius draw an arc cutting LK in two 
points as P and P'. Draw OP and OP' ; then A is any angle coterminal 
with Z XOP or Z XOP'. For draw PM JL X'X and P'Jlf ± X'X ; then 
esc XOP = 5/(- 3) = esc XOP'. 

om = - V5 2 - 3 2 = - 4, oar = 4. 

.-. sin .4 = - 3/5, esc ^4 = - 5/3 ; 

cos J- = T 4/5, sec A—^ 5/4 ; 

tan .4 = ± 3/4, cot A = ± 4/3. 

13. Express each of the trigonometric ratios of A in terms of sin .4. 

If sin A is positive, -4 is in the first or pi y ^p 

the second quadrant. 

In fig. 5, let OP = 1. 

Then sin ^4 is the measure of MP 
or M'P / . 

Whence .MP = ilf'P' = sin A, 

OM = Vl -sin 2 A, 

OM' = - Vl-sin 2 ^4. 




X'M 



secA = ± 1/Vl — sin 2 ^4 ; 



Hence cos J. = ± vl — sin 2 A , 
tan J. = ± sin J./Vl — & 
If sin A is negative, A is in the third or fourth quadrant, and 



A, cotAz=±Vl — sin 2 A /sin ^4. 



cos ^4 = =F Vl - sin 2 ^4, 



sec J. = T 1/Vl - sin 2 A. 



EXERCISE VII 

1. Given sin A = — 2/3, compute the other trigonometric ratios of A. 
Since sin A is — , A is in the third or fourth quadrant, 
sin A = - 2/3. 
.-. esc A = - 3/2. by [1] 

by [4] 



cos A = =F Vl — sin 2 A 



= T Vl - 4/9 = T V&/3. 

sec ^4 = 1/cos A = 1/(T V5/3) = T 3/^/5. by [1] 

tan J. = sin ^4/cos A = (- 2/3)/( =F V5/3) = ± 2/^/5. by [2] 

cot A = 1/tan J. = ± V5/2. by [1] 



28 PLANE TRIGONOMETRY 

2. Given cos .A — 1/3, compute the other trigonometric ratios of A. 
Since cos A is + , A is in the first or fourth quadrant. 

cos A = 1/3. 
.-. sec A = 3. 

sin A = ± Vl - cos 2 ^. = ± Vl - 1/9 

= ± V8/3 = ± 2 V2/3. 
.-. esc A = ± 3/(2 V2). 

tan J. = sin 4/cos A = ±2 y/2. 

.-. cot 4 = ± 1/(2 V2). 

3. Given sin A = 0.2, compute the other trigonometric ratios of A. 
Since sin A is + , A is in the first or second quadrant. 

sin A = 0.2. 
.-. esc A = 1/0.2 = 5. 

cos A = ± Vl - sin 2 J. = ± Vl - (1/5) 2 
= ± V24/5 = ± 2 V6/5. 
.-. sec J. = ± 5/(2 V6). 

tan A = sin J. /cos A = ± 1/(2 V^)- 
.;. cot A = ± 2 V6. 

4. Given cos A = — 3/4, compute the other ratios of A. 
Since cos A is — , A is in the second or third quadrant. 

cos A = - 3/4. 
.-. sec A = - 4/3. 

sin A = ± Vl - 9/16 = ± V7/4. 
.-. esc A = ± 4/V7. 

tan J. = sin A/cos A : = =p W3. 
• .-. cot A = t 3/V7. 

5. Given tan .A = — 4/3, compute the other ratios of A. 
Since tan A is — , A is in the second or fourth quadrant. 

tan A = - 4/3. 
.-. cot A = - 3/4. 

sec J. = T Vl + tan 2 ^4 = =F 5/3. 
.-. cos A = =f 3/5. 

sin J. = ± Vl - cos 2 ^4 = ± 4/5. 
.-. esc A = ± 5/4. 



EXERCISE VII 29 

6. Given cot A = — 2, compute the other ratios of A. 
Since cot J. is — , A is in the second or fourth quadrant. 

cot A = - 2. 
.-. tan A = -1/2. 

esc A = ± Vl + cot 2 A = ± V5- 
.-. sin A = ± 1/V5. 

cos ^1 = T Vl - sin 2 A = =F 2/V&. 
.% sec J. = ^ V 6 / 2 - 

7. Given cot A = 3/2, compute the other ratios of A. 
Since cot A is + , A is in the first or third quadrant. 

cot A = 3/2. 
.-. tan A = 2/3. 

esc A = ± Vl + cot 2 J. = ± V13/2. 
.-. sin ^1 = ± 2 A/13. 

cos J. = ± Vl-sin 2 ^L = ± 3/V13. 
.-. sec A = ± V13/3. 

8. Given tan .4 = 2.5, compute the other ratios of A. 
Since tan A is + , A is in the first or third quadrant. 

tan A = 5/2. 
.-. cot A = 2/5. 

sec A = ± Vl + tan 2 A = ± V29/2. 
.-. cos A = ± 2/V29. 

sin A = ± Vl - cos 2 .4 = ± 5/V29. 
.-. esc A = ± V29/5. 

9. Given esc A — — V3, compute the other ratios of A. 
Since esc -4 is — , A is in the third or fourth quadrant. 

esc A = — V3. 
.-. sin A = - 1/V3. 

cos A = =p Vl - sin 2 ^i = T V2/3. 
.-. sec J. = T V372. 

cot A = ± Vcsc 2 ^. - 1 = ± V2. 
.-. tan A = ± 1/V2 = ± V2/2. 



30 PLANE TRIGONOMETRY 

10. Given sec ^4 = 4, compute the other ratios of A. 
Since sec A is + , A is in the first or fourth quadrant. 

sec A = 4. 
.-. cos A = 1/4. 

sin A = ± Vl - cos 2 A = ± V15/4. 
.-. esc A = ± 4/V15. 

tan A = sin J./cos A = ± -y/15. 
.-. cot 4 = ± 1/V15 = ± V15/15. 

11. Given tan A = - y7, compute the other ratios of A. 
Since tan A is — , ^4 is in the second or fourth quadrant. 

tan A = — y/7. 
.-. cot ^4 = - 1 /V7 = - V 7/7. 

sec J. = T Vl + tan 2 J. = =f V8 = T 2 V2. 
.-. cos 4 = T 1 /(2V2) = T V2/4. 

sin .4 = ± Vl - cos 2 A = ± V14/4. 
.-. esc A = ± 4/V14. 

12. Given cos ^4 = m/c, compute the other ratios of A. 

If m and c are both positive, A is in the first or fourth quadrant. 

cos A = m/c. 
.-. sec A = c/m. 

sin A = ± V 1 — cos? A = ± Vl - (m/c) 2 = ± ^Ttf^wt/c. 
.-. esc A = ± c/Vc? — m 2 . 

tan J. = sin ^4 / cos ^4 = ± Vc 2 — m 2 /m. 
.-. cot ^4 = ± wi/Vc 2 — m 2 . 

If m is negative and c positive, A is in the second or third quadrant, 
but all the values given above are still correct. 

13. Express each of the trigonometric ratios of A in terms of cos A. 
If cos A is + , A is in the first or fourth quadrant. 

sec A = 1/c os A. 

sin A = ± Vl — cos 2 ~X 
.-. esc A = ± 1 /Vl — cos 2 v4. 

tan A = ± Vl — co s 2 ~^4/cos ^4. 
.-. cot J. = ± cos ^4/Vl - cos 2 J.. 

If cos A is — , A is in the second or third quadrant, but all the relations 
given above still hold true. 



EXERCISE VII 31 

14. Express each of the trigonometric ratios of A in terms of tan A. 
If tan A is + , A is in the first or third quadrant. 

cot A = 1/t an A. 

sec A = ± Vl + tan 2 ^.. 
.-. cos A = ± 1/Vl + tan 2 A. 



esc A = ± Vl + l/ tan 2 ^ = ± Vtan 2 A + 1/tan A. 
.-. sin -4. ="± tan A /Vtan 2 A + 1. 
If tan A is — , A is in the second or fourth quadrant, and 
esc A = T Vtan 2 ^4 +T/tan^4, 
sin A = ^ tan ^4 /Vtan 2 ^4 + 1. 

15. Express each of the trigonometric ratios of A in terms of cot A. 
If cot A is + , A is in the first or third quadrant. 

tan A = 1/c ot^L. 

esc A = ± Vl + cot 2 Z. 
.-. sin A = ± 1/Vl + cot 2 ^4. 



sec ^4 = ± Vl + l/cot 2 A = ± vcatM+T/cot-4. 
.-. cos A = ± cot jyVcot 2 ^ + 1. 
If cot A is — , A is in the second or fourth quadrant, but all the rela- 
tions given above still hold true. 

16. Express each of the trigonometric ratios of A in terms of sec A. 
If sec A is + , A is in the first or fourth quadrant. 

cos A = 1/sec^L 

tan A = ± Vse c 2 J. — 1. 
.-. cot^EE±l/Vsec 2 ^l -1. 

sin A = ± Vl - l/ sec^A = ± Vsec 2 ^4 - 1/sec A. 
.-. esc A = ± sec J. /Vsec 2 ^4. — 1. 
If sec A is — , A is in the second or third quadrant, and 

sin A = =F Vsec 2 A -^T/sec ^L , 

esc .4 = ^ sec J. /Vsec 2 A — 1. 

17. Express each of the trigonometric ratios of A in terms of esc A. 
If esc ^4 is + , A is in the first or second quadrant. 

sin A =l/csc^4. 
cot A ~ ± Vcs c 2 ^4 - ~1. 
.-. tan A = ± 1 /Vcsc 2 ^4 - 1 . 

cos A = ± Vl - 1/ cscM" = ± Vcsc 2 A - 1/csc A. 
.-. sec A = ± esc J. /Vcsc 2 A — I. 
If esc J. is - , ^4 is in the third or fourth quadrant, but all the relations 
given above still hold true. 



32 PLANE TRIGONOMETKY 

EXERCISE VIII 

1. Prove cos A tan A = sin A. 

By [2], tan A = sin A /cos A . .-. cos A tdmA = sinA. 

2. Prove sin A sec A = tan A. 

By [2] , tan A = sin A /cos .4 = sin 4/(l/sec A) = sin J. sec A. 

3. Prove cos A esc A = cot A. 

By [3], cot A = cos J. /sin J. = cos ^./(1/csc A) = cos A esc 4.. 

4. Prove sin A cot A = cos J.. 

By [3], cot A = cos A/sin A. .-. sinAcotA = cosA. 

5. Prove cos 2 4 - sin 2 A = 1-2 sin 2 ^4. (1) 
By [4], cos 2 A + sin' 2 A = 1. (2) 
Subtracting 2 sin 2 4 from each member of (2), we obtain (1). 

6. Prove cos 2 A — r sin 2 A = 2 cos 2 ^4 — 1. 

By [4], cos 2 ^4 - sin 2 A = cos 2 A - (1 - cos 2 A) = 2 cos 2 4 - 1. 

_ _ sini 1-cosi 

7. Prove = (1) 

1 + cos A sin A 

By [4], sin 2 J. = 1 - cos 2 A = (1 -cos A) (1 + cos J.). (2) 

Dividing the members of (2) by sin J. (1 + cos J.), we obtain (1). 

_ _. 1 + sin A cos A 

8. Prove — = (1) 

cos A 1 — sin A 

By [4], cos 2 J. = l- sin 2 ^4 = (l + sin4)(l-sin^). (2) 

Dividing the members of (2) by cos^l (1 — sin A), we obtain (1). 

■ _, sec A + 1 tan A ... 

9. Prove = (1) 

tan A sec A — 1 

By [5], tan 2 A = sec 2 A - 1 = (sec A + l) (sec A - 1). (2) 

Dividing the members of (2) by tan A (sec A — 1), we obtain (1). 

10. Prove sec ^4 + tan 4 = l/(sec A — tan A). (1) 
By [5], lEEsec 2 A-tart i A = (secA + ta,nA)(secA-timA). (2) 
Dividing the members of (2) by sec ^4 — tan A, we obtain (1). 

11. Prove (1 + tan 2 A) cos 2 A = 1. 

(1 + tan 2 4) cos 2 A = sec 2 A cos 2 A = 1. by [5], [1] 



EXERCISE VIII 33 

12. Prove (1 + cot 2 ^l) sin 2 ^. = 1. 

(1 + cot 2 ^1) sin 2 J. = csc 2 ^4 sinM = 1. by [6], [1] 

13. Prove sin 2 A + sin 2 A tan 2 A = tan 2 A. 

sin 2 A (1 + tan 2 A) = sin 2 J. sec 2 A = sin 2 ^4/cos 2 A = tan 2 J.. 

14. Prove (esc 2 A — 1) sin 2 ^L = cos 2 J.. 

(esc 2 J. - 1) sin 2 J. = cot 2 A sin 2 A = cos 2 A. 

15. Prove cos 4 J. - sin 4 A + 1 = 2 cos 2 ^4. 

cos 4 A - sin 4 J. + 1 = (cos 2 ^4 + sin 2 J.) (cos 2 A - sin 2 A) + 1 
= cos 2 J. + (1 - sin 2 A) = 2 cos 2 A. 

16. Prove tan 2 ^4/(1 + tan 2 J.) = sin 2 A. 

tan 2 ^/(l + tan 2 A) = tan 2 A /sec 2 A = tan 2 A cos 2 A = sin 2 A. 



17. Prove Vl — sin 2 A /sin A = cosyl/Vl — cos 2 ^.. 

(1 - sin 2 ^4) /sin 2 ^4 = cos 2 ^4/(1 - cos 2 A). 
.-. Vl — sin 2 A /sin A = cos JL/Vl — cos 2 A. 

18. Prove cot 2 A — cos 2 A = cot 2 A cos 2 A. 

cot 2 A - cos 2 A = cos 2 A (1/sin 2 A - 1) ee (esc 2 A - 1) cos 2 A 

= cot 2 ^4 COS 2 J.. 

19. Prove sec 2 A + csc 2 ^4 = sec 2 J. esc 2 A. 

1 1 



sec 2 A + esc 2 A = 



cos 2 A sin 2 .4 
sin 2 A + cos 2 vl 
sin 2 ^4 cos 2 J. 
1 
sin 2 A cos 2 A 
1 1 



sin 2 A cos 2 ^4 
= sec 2 A esc 2 A. 

20. Prove tan A + cot A = sec A esc A. 

a , 4. a sin J. cos J. 

tan A + cot A = 1 

cos A sin A 

sin 2 A + cos 2 A 



cos A sin A 
1 1 



cos A sin A 
= sec J. esc ^4. 



34 PLANE TRIGONOMETRY 

„ cot A cos A cot A — cos A 

21. Prove = — ■ (1) 

cot A + cos A cot A cos A 

cot A cos A cos 2 .4/sin A 



Also 



cot ^4 + cos A cos -4 /sin A -\- cos -4 

= cos 2 J./[cos A (1 + sin 4.)] 
= (1 - sin 2 ^4)/[cos^4(l + sin^L)] 
ee(1 -sin.4)/cos^4. (2) 

cot A — cos A _ cos A /sin J. — cos .4 
cot 4. cos A cos 2 .4/sin .4 

_ cos .4(1 — sin ^4) 

cos 2 ^4 
= (1- sin A) /cos A. (3) 

From (2) and (3) we obtain (1). 

22. Prove tan A + cot A = (sec 2 .A -f esc 2 ^4) /(sec A csc^4). 
(sec 2 v4 -l- csc 2 .4)/(sec A esc ^4) = sec J./csc A + esc ^4/sec A 

ee sin 4L/cos J. + cos A /sin v4 
= tan .4 + cot .4. 

23. Prove 1/Vsec 2 ^4 - 1 = Vcsc 2 ' A — 1. 

1/Vsec 2 ^4 - 1 = 1/tan .4 

= c ot J. 

= Vcsc 2 ^. - 1. 

_": _ 1 + tan 2 ^4 sin 2 A esc A . 

24. Prove = ; = cos J.. 

1 + cot 2 A cos 2 A cot A + tan A 

1 + tan 2 ^4 _ sec 2 A 

1 + cot 2 A ~ esc 2 J. 

sin 2 ^4 



cos 2 J. 



// sin 2 .4\ 
(cos A H ) 
\ cos A J 

= cos.4/(cos 2 .4 4- sin 2 J.) 

= cos^l. 

ni _ _ cos J. sin A . 

25. Prove 1 Esini + cos A. 

1 — tan A 1 — cot A 

cos A sin A _ cos 2 A sin 2 .4 



1 — tan A 1 — cot A cos A — sin A sin .4 — cos A 
_ cos 2 -4 — sin 2 A 
cos A — sin A 
= cos J. + sin .4. 



EXERCISE VIII 35 

26 Prove sin 3 .4 cos A + cos 3 .4 sin A = sin A cos .4. 

sin 3 A cos A + cos 3 A sin A = sin A cos J. (sin 2 A + cos 2 J.) 
= sin_<4. cos A. 

27. Prove sin 2 ^4. cos 2 ^4 + cos 4 .J. = 1 — sin 2 A, 

(sin 2 .4 + cos 2 A) cos 2 A = cos 2 J. 

= 1- sin 2 ^4. 

«« -r, /l — sin 4 . . 

28. Prove + / = sec ^4 — tan A. 

sin A 



V I - sin A _ 1 (1- sin Ay 2 
1 + sinJ. - \ l-sin 2 J. 
_ 1 — sin A 

cos A 
= sec A — tan A. 

_ _. sin A 1 + cos A . 

29. Prove 1 = 2 csc^.. 

1 + cos .4 sin A 

sin A 1 + cos A _ sin 2 J. + (1 + cos^4) 2 

1 + cos^l sin A sin A (1 + cos A) 

_ sin 2 A + cos 2 A + 1 + 2 cos .4 

sin A(l + cos ^4.) 

_ 2(1 + cos A) 



sin ^4.(1 + cosyl) 
2 esc A. 



30. Prove l/(cot A + tan A) = sin A cos A. 
l/(cot A + tan ^4) = 



cos .4/sin A + sin ^4,/cos A 
sin J. cos ^4 



cos' 2 A + sin 2 ^4. 
31. Prove l/(sec A - tan A) = sec J. + tan A. 



sin A cos A. 



l/(sec J. - tan A) = 1 //— si ^\ 

/ \cos J. COS-4./ 



/ Vcos A 


cos A, 


cos ^4/(1 - 


sin A) 


cos ^4(1 + 


sin J.) 


1 — sin 1 


*A 


1 sin A 

r + r 



cos A cos A 
= sec A + tan A. 



36 PLANE TRIGONOMETRY 

1 — tan A cot A — 1 



32. Prove 



1 + tan A cot A + 1 

1 — tan A cos A — sin A 



1 -f tan A cos ^L -f sin A 
_ cos J. /sin A — 1 
~~ cos A/s'm A + 1 
cot J. — 1 



cot J. + 1 



33. Prove = cos 2 A — sin 2 J.. 

1 + tan 2 A 

1 - tan 2 A cos 2 A - sin 2 A 



1 + tan 2 J. cos 2 J. + sin 2 A 
= cos 2 A — sin 2 .4. 

34. Prove esc A/ '(cot -4 + tan J.) = cos A . 

esc J. 1/sin A 



cot ,4 + tan A cos ^4/sin ^4 + sin A /cos ^4 
cos .4 



cos 2 ^4 + sin 2 ^4 
cos J.. 



35. Prove esc 4 A (1 - cos 4 ^l) - 2 cot 2 A = 1. 

esc 4 ^4 (1 - cos 4 J.) - 2 cot 2 J. = csc 4 J. sin 2 J.(l + cosM) -2 cotM 
= esc 2 A (1 + cos 2 ^4) - 2 cot 2 J. 
= csc 2 v4 + cot 2 A - 2 cot 2 A 
= esc 2 A — cot 2 A = 1. 



EXERCISE IX 

Express each of the following trigonometric ratios in terms of the ratio 
of some positive acute angle less than 45°. 

1. sin 168° = cos 78° = sin (90° - 78°) = sin 12°. §§ 29, 30 
Or sin 168° = sin (180° - 168°) = sin 12°. §30 

2. tan 137° = - cot 47° = - tan (90° - 47°) = - tan 43°. §§ 29, 30 
Or tan 137° = - tan (180° - 137°) = - tan 43°. 

3. cos 287° = - sin 197° = - cos 107° = sin 17°. § 29 
Or cos 287° = cos (3 -90° + 17°) = sin 17°. §31 



EXERCISE IX 37 

4. sin 834° = sin (2 • 360° + 114°) = sin 114° = cos 24°. 
Or sin 834° = sin (9 • 90° + 24°) = cos 24°. 

5. sin ( - 65°) = - sin 65° = - cos 25°. 

6. cps (- 84°) = cos 84° = sin 6°. 

7. tan (- 246°) = tan 114° = - cot 24°. 

8. cos ( - 428°) = cos (- 68°) = cos 68° = sin 22°. j 

9. cos 1410° = cos (15 • 90° + 60°) = sin 60° = cos 30°. 

10. tan 1145° = tan (12 • 90° + 65°) = tan 65° = cot 25°. 
Or tan 1145° = tan (3 ■ 360° + 65°) = tan 65° = cot 25°. 

11. cot 1054° = cot (11 ■ 90° + 64°) = - tan 64° = - cot 26°. 

12. sec 1327° = sec (14 • 90° + 67°) = - sec 67° = - esc 23°. 

13. esc 756° = esc (8 ■ 90° + 36°) = esc 36°. 
Or esc 756° = esc (2 • 360° + 36°) = esc 36°. 

14. tan ( - 196° 54') = - tan 196° 54' = - tan (2 • 90° + 16° 54') § 28 

= - tan 16° 54'. § 31 

15. cot ( - 236° 21') = cot 123° 39' = - tan 33° 39'. §§ 21, 29 

16. Prove sin 420° cos 390° + cos (- 300°) • sin (- 330°) = 1. 

sin 420° = sin (360° + 60°) = sin 60° = V3/2. 

cos 390° = cos (360° + 30°) = cos 30° = V3/2. 
cos(- 300°) = cos 60° = 1/2. 
sm(-330°) = sin30°=:l/2. 

.-. sin 420° cos 390° + cos (- 300°) sin (- 330°) = X?.^? + i.l = i. 
v ; v '2-222 

17. Prove cos 570° sin 510° - sin 330° cos 390° = 0. 

cos 570° = cos (6 • 90° + 30°) = - cos 30° = - y/S/2. 
sin 510° = sin (5 • 90° + 60°) = cos 60° = 1/2. 
sin 330° = sin (- 30°) = - sin 30° = - 1/2. 
cos 390° = cos (360° + 30°) = cos 30° = V3/2. 

.-. eos 570° sin 510° - sin 330° cos 390° = -— ■ - + - ■ — = 0. 

2 2 2 2 



§21 


§ 


i31 


§§28, 


30 


§§28, 


30 


§§21, 


30 


21, 28, 


30 


§§31, 


30 


§§31, 


30 


§§21, 


30 


§§31, 


30 


§§31, 


30 


§31 


§21 



38 PLANE TRIGONOMETRY 

EXERCISE X 

1. State in words identities [7] and [8], as generalized in § 35. 

The sine of the sum of "1 _ J sin first • cos second 
any two angles J ~ I + cos first • sin second. 



The cosine of the sw?w"j _ ( cos first • cos second 
of any two angles J — \ — sin first • sin second. 



2. Putting 75° = 30° + 45° and using [7], find sin 75°. 

sin 75° = sin (30° + 45°) 

= sin 30° cos 45° + cos 30° sin 45° by [7] 

1 V2 V3 V2 _ V2 + V6 
~2* 2 + 2 ' 2 4 

3. Putting 75° = 30° + 45° and using [8], find cos 75°. 

cos 75° = cos (30° + 45°) 

= cos 30° cos 45° - sin 30° sin 45° 
_ V3 V2 _ 1 V2 _ V6 -V2 
__ 2 2~~ %'~Y ~ 4 

4. Putting 15° = 45° + (- 30°), find sin 15° and cos 15°. 

sin 15° = sin [45° + (-30°)] 

= sin 45° cos ( - 30°) + cos 45° sin ( - 30°) 

H) 



V2 V3 V2 / 1\ V6 - V2 
~2 2~ + _ ¥ 



cosl5° = cos[45° + (-30°)] 

= cos 45° cos ( - 30°) - sin 45° sin ( - 30°) 
_ V2 V3 V2 / 1\ _ V6 + V2 

_ ~2 2 



(-D- 



5. Putting 15° = 60° + (- 45°), find sin 15° and cos 15°. 

sin 15° = sin [60° + (-45°)] 

= sin 60° cos ( - 45°) + cos 60° sin ( - 45°) 
_ V3 V2 1 / V2\ V6 - V2 
Z 2"2 + 2\ 2 / 4 

cos 15° = cos [60° + (- 45°)] 

= cos 60° cos ( - 45°) - sin 60° sin ( - 45°) 
_1 V2 -y/3 / V2\ _ V2 + V6 
_ 2'~2 2~\ ~2~/~ 4 



EXEKCISE X 39 



6. Putting 90° = 60° + 30°, find sin 90° and cos 90°. 

sin 90° = sin (60° + 30°) 

= sin 60° cos 30° + cos 60° sin 30° 

V3 V3 11 3 1 , 
= -^— • — + = - + - = 1. 

2 2 2 2 4 4 

cos 90° = cos (60° + 30°) 

= cos 60° cos 30° - sin 60° sin 30° 

= 1 . V3 _ V3 1 = Q 
2' 2 "" 2 2 



7. Putting 0° = 45° + (- 45°), find sin 0° and cos 0°. 



sin0 = sin[45°+(-45 )] 

= sin 45° cos (- 45°) + cos 45° sin (- 45°) 

2 ' 2 



- 45°)] 

- 45°) - sin 

-f(-f) 



cos 0° = cos [45° + (-45°)] 

= cos 45° cos ( - 45°) - sin 45° sin ( - 45°) 
_ V2 V2 
~~2 2~ 



8. Given sin A = 2/5 and cos B = 1/3, A and B being positive acute 
angles ; find the values of sin (A + B) and cos {A + B). 

sin (A + B) = sin A cos B + cos A sin B. [7] 

cos (A + B) = cos A cos B — sin A sin B. [8] 

cosJ.= Vl~-sin 2 ^l = V l - 4/2 5 = V21/5. 
sin B = Vl - cos 2 B = Vl - 1/9 = 2 V2/3. 

Substituting for sin ^4, cos ^4, sin -B, cosJB in [7] and [8] their values, 
we obtain 



cos {A + 5) 



5 3 5 3 15 

•v/21 12 2 V2 _ V21 - 4 V2 
5 '3 5 3 15 



9. Given sin A = 2/3 and cos B = 1/4, A and 5 being positive acute 
angles ; find the values of sin (A + B) and cos(^L + B). 

cos A = Vl-sin^^l = Vl _ 4/9 = vV3. 
sin 5 = Vl - cos 2 £ = Vl -1/16 = VI 5/4. 



40 



PLANE TRIGONOMETRY 



Substituting in [7] and [8] these values for sin A, cos A, sinJ5, cos J?, 
we obtain 

2 1 ., V5 VI 5 _ 2 + 5 V3 

4 



sin (A + B) = 
cos (J. + 5) 



3 4 3 4 12 

V5 1 _ 2 V15 _ V5 - 2 V15 

~3~'i 3~i~ - 12 



10. Putting 90° + A for J. in [7], deduce [8]. 

Putting 90° + A for A in [7], we obtain 

sin [90° + (A + #)] = sin (90° + 4) cos B + cos (90° + A) sin J5. 
.-. cos (A + B) = cos J. cos B — sin A sin I?. 

11. Putting 90° + A for A in [8], deduce [7]. 

Putting 90° + .A for A in [8], we obtain. 

cos [90° + (A + B)] = cos (90° + ^1) cos jB - sin (90° + ^1) sin 5. 
.-. — sin ( A + B) = — sin ^i cos I? — cos A sin J5, 
or sin {A + 1?) •= sin A cos .B + cos A sin 2?. 

12. Prove [7] and [8], using trigonometric lines, A and B being in 
the first quadrant. 




i 


• 
P 






D 


K \ Tn. 


I 


/•* 


\ / ' 


JS 


;*. 



M 



Fig. 6 



In each figure let Z XOE = A, 



and ZROC = B. 

On OC take OP = + 1. 

Then JfP = sin (A + B), 

ON = cos B, NP = sin B. 
.: DP = NP cos DPN = sin J? cos A. 
QN = ON sin A = cos B sin J.. 
.-. sin (A + B) = QN + DP = sin AcosB -\- cos ^1 sin B. 



EXERCISE XI 41 



EXERCISE XI 

1. State in words identities [9] and [10]. 

The sine of the difference \ _ ( sin first • cos second 
of any two angles J ~ i.~ cos first ■ sin second. 

The cosine of the difference ~) _ f cos first • cos second 
of any two angles J ~~ I + sm fi rs ^ ' sm s & c ond. 

2. Putting 15° = 45° - 30°, find sin 15° by [9] and cos 15° by [10]. 
By [9] , sin 15° = sin (45° - 30°) = sin 45° cos 30° - cos 45° sin 30°. 

_ V2 V3 _ V2 1 _ V6-V2 
~~2 2 2~'2 ~ 4~~ 

By [10] , cos 15° = cos (45° - 30°) = cos 45° cos 30° + sin 45° sin 30° 
_ V2 V3 V2 1 _ V6 + V2 
~ 2 ' 2 2 *2 _ 4 

3. Putting 15° = 60° - 45°, find the sine of 15° and the cosine of 15°. 
By [9], sin 15° = sin (60° - 45°) = sin 60° cos 45° - cos 60° sin 45° 

_ V3 V2_l V2 _ V6 - V2 

"2*2 2' 2 ~ 4 

By [10], cos 15° = cos (60° -45°) = cos 60° cos 45° + sin 60° sin 45° 

_ 1 V2 V3 V2 _ V2 + V6 

~ 2 ' T~ + ~2 2~ ~~ 4~~ 

4. A and B being positive acute angles, find the values of sin {A — B) 
and cos {A — B), having given sin A = 1/4, sinJ3 = 1/3. 

cos A = V l - sin 2 ^l = Vl - 1/1 6 = V15/4. 
cos B= Vl - sin 2 # = Vl -1/9 = 2 V2/3. 

• ,A 1>S 'A -O ■ T> A 1 2V2 1 V15 

sm (A — B)~ sin A cos B — smB cos A = - ■ — ■ — — 

= (2 V2 - V15)/12. 6 

, a ™ , • , • t, V15 2 V2 11 
cos ( A — B) = cos AcosB + sm ^i sin I? = \- -• - 

= (2 V30 + 1)/12. 4 3 4 3 

5. A and jB being positive acute angles, find the values of sin (A — B) 
and cos(^i — B), having given cos A = 2/3, cosi? = 3/4. 

sin A = Vl - cos 2 A = V l - 4/9 = V5/3. 
sin B = Vl - cos 2 B = Vl - 9/16 = V7/4. 

sin ( J. — B) = sin J. cos B — cos J. sin B = — ■^— 

= (3V5-2V7)/12. 3 4 3 4 

cos (A — B) = cos A cos 5 + sin A sin J5 = h — • ^— - 

= (6+V35)/12. 3 4 3 4 



42 PLANE TRIGONOMETRY 

6. Prove sin (A + B) sin (A - B) = sin 2 A - sin 2 B. 
sin (A + B) sin (J. - B) 

= sin 2 ;! cos 2 5 - cos 2 J. sin 2 7? + (sin 2 J. sin 2 £ - sin 2 A sin 2 B) 
= sin 2 A (cos 2 5 + sin 2 J3) - sin 2 5 (cos 2 A + sin 2 ^i) 
= sin 2 ^.-sin 2 £. 

Observe that sin 2 A sin 2 B — sin 2 J. sin 2 1? is added above as one form 
of zero. 

7. Prove cos (A + B) cos ( A - B) = cos 2 J. - sin 2 5. 
cos (J. + B) cos (J. - JS) 

= cos 2 J. cos 2 B - sin 2 A sin 2 5 + (cos 2 A sin 2 B - cos 2 A sin 2 B) 
= cos 2 A (cos 2 B + sin 2 B) - sin 2 # (sin 2 A + cos 2 ;!) 
= cos 2 ^i — sin 2 jB. 

8. Prove sin (A + B) cos B — cos (J. + B) sin 5 = sin A. 
sin (J. + J5) cos B — cos (J. + B) sin i> 

= sin A cos 2 B + sin B cos J3 cos ;! — cos J. cos BsinB -\- sin w4 sin 2 B 
= sin J. (cos 2 B + sin 2 5) 
= sin A. 

9. Prove sin (J. + B) + cos (J. - B) = (sin J. + cos A) (sin 5 + cos B). 
sin (;! + B) + cos (J. - B) 

= sin ^! cos B + cos ^! sin B + cos ^1 cos J5 + sin J. sin B 
= sin A (sin E + cos B) + cos ;! (sin B -f- cos 5) 
= (sin A + cos J.) (sin B + cos 5). 

10. Prove sin A cos (B — C) - sin B cos (;! — C) = sin (J. — B) cos C. 

sin A cos (5 - C) - sin 5 cos (A - C) 

= sin A cos .B cos C+sinA sin J3 sin C — sin B cos J. cos C — sinB sin J. sin C 
= sin (J. - B) cos C. 

11. Prove tan A + tan J5 = sin (A + l?)/(cos A cos 5). 

sin M + B) sin ^ cos B + cos J. sin B _, 

i '- = = tan A + tan B. 

cos A cos 5 cos A cos 5 

12. Prove cot B — cotA = sin (J. — B)/(sin A sin J5). 

sin (A — B) sin J. cos B — cos ;! sin B . 

i L = = cot B — cot A. 

sin A sin B sin A sin 5 



EXERCISE XII 



43 




13. Prove [9] and [10] geometrically, using trigonometric lines, when 
A, B, and A — B are in the first quadrant. 

Let XOB and ROC be any two acute angles, 
ZROC being negative and ZXOC being positive. 

Then Z XOC = Z XOR + ZROC. 

Let A denote any angle coterminal with XOR, 
and — B any angle coterminal with ROC. 

Then A + (— B), or A — B, will be coterminal 
with XOC. 

Take OP equal to +1. 

Draw PM± OX, PN JL OR, NQ ± OX, and PD _L QN. 

sin ( — B) = — sin B, 
cos ( — B) — cos B, 
MP=QN- DN 
ON ■ sin XOR = cos B sin A, 
( - NP) cos DNP = sin 5 cos A. 
QN — DN = sin A cos 5 — cos A sin 5. 
ON cos XOB = cos B cos ^4, 
( - NP) sin ZWP = sin BsmA. 
cos (A - 5) = OM = OQ+DP = cos A cos B + sin ^i sin B. 



Now 




NP 

ON 


and 


sin (A 


-B) 


Also, 




QN 


and 




DN 




.-. sin (A 


-B) 


Again, 




OQ 


and 




DP 



EXERCISE XII 



1. State in words identities [11] and [12]. 

the sum of their tangents 
— product of their tangents 



The tangent of the sum of any 1 
two angles j 



of any two angles 



I j me t 



product of their tangents 



2. Putting 75° = 45° + 30°, find 3. Putting 15° = 60° - 45°, find 



tan 75° by [11]. 

tan 75° = tan (45° + 30°) 

tan 45° + tan 30° 
~ 1 - tan 45° tan 30 c 
^ 14^3/3 
1-V3/3 
= V3 + 1 
V3-1* 



tan 15° by [12]. 

tan 15° = tan (60° - 45°) 

tan 60° - tan 45° 

~~ 1 4- tan 60° tan 45° 

- V 3 -! 

~ V3 + 1 ' 



44 PLANE TRIGONOMETRY 

4. If tan J. = -1/2 and tan 5=3, tan A — tan B 
find tan (A + B) and tan (A—B). an ^ ~~ ' = l + tan A tan B 

. A , ■ tan J. + tan 5 -2+31 

tan ( J. + £) = = — — = - • 

v l-tan.4tan£ 1 + 6 7 

1/2-4-3 6l Prove 

-^vMi^ tan(45Q + ^)^ 1 + tan ^. 

1 + d / 2 V ' 1-tanJ. 

ten (.1-3) = ** A -*** B tan(15 o,iw tan 45o + tan ^ 

1 + tan^tanB tan (45 + ^) = 2 _ tan45 o tan ^ 

= - 1/2 - 3 _ _ 1 + tan J. 

1-3/2 ' ~l-tanJ.' 
7. Prove 

5. If tan A =—2 and tan B=— 3, _l_ tan^i 
find tan (J. + B) and tan (A - B). tan ( 45 ° ~ A ) = i + tan^L * 

t.iTW.4 1 Tft- tan ^ + tan - B ,.„ ,, tan45°-tanA 

tan {A + ±>) = - tan (45° - A) = 

1 -tan -4 tan B v ' 1 + tan 45° tan .4 

_ -2 - 3 _ = l-tan^l 

1 - 6 ~ ' ~ 1 + tan^.' 

cot A cot B — 1 



8. Prove cot (X + B) = 
cot (^L + B) = 



cotB + cot .A 
cos (A + J5) 



sin (A + 5) 
_ (cos A cos B — sin A sin i?) h- (sin A sin 5) 
— (sin A cos .B + cos A sin i?) h- (sin A sin B) 
_ cot A cot .B — 1 
cotS + cot^L 

« -r, w . „. coticotB + 1 * 

9. Prove cot (X - 5) = — . (1) 

cot J5 - cot A 

Substituting — B f or B in the relation in example 8, we obtain 

cot( 4-*) = ===t=)^ (2) 

v ' cot(-5) + cotJ. 

Substituting — cot B for cot(— B) in (2), we obtain (1). 
10. Prove identity [12] by dividing [9] by [10]. 
Dividing [9] by [10], we obtain 

sin A cos B — cos A sin B 



tan (A - B) 



cos A cos B + sin A sin 5 
tan A - tan 5 



1 + tan A tan B 



EXERCISE XIII 45 

11. Prove the identities in examples 8 and 9 by taking the reciprocals 
of the members of [11] and [12] respectively. 
Taking the reciprocal of the members of [11], we obtain 
1 — tan A tan B cot A cot B 



cot {A + B) = 



tan A + tan B cot A cot B 
cot A cot J? — 1 



cot B + cot A 
Taking the reciprocal of the members of [12], we obtain 
1 + tan A tan B cot A cot B 



cot (A -B) 



tan A — tan B cot A cot B 
cot A cotB + 1 
cot B — cot A 



12. Eind tan (A + B) and tan (A — B) in terms of cot A and cot B. 
tan A + tan B cot J. cot B 



tan (4 + B) = 



tan (J. - B) 



1 — tan .A tan B cot J. cot B 

cotB + cot ^4 
cot A cotB — 1 

tan J. — tan B cot ^4 cot 5 



1 + tan A tan B cot -4 cot 2> 
cot 2? — cot A 



cot A cotB + 1 



13. Find cot {A + B) and cot (A — B) in terms of tan^l and tan B. 
Taking the reciprocals of the members of [11], we obtain 

* / a , t,\ 1 — tan A tan A 

cot (A + B) = 

tan A + tan J5 

Taking the reciprocal of the members of [12], we obtain 

1 + tan A tan B 



cot (A - B) 



tan A — tan B 



EXERCISE XIII 

State in words identities [13], [14], and [15]. 
sin twice an angle = 2 sin angle • cos angle. 
cos twice, an angle = (cos angle) 2 — (sin angle) 2 

= 1 — 2 • (sin angle) 2 = 2 (cos angle) 2 — 1. 
tan Zuuce an angle = 2 • tan cma£e/[l — (tan angle) 2 ]. 



46 PLANE TRIGONOMETRY 

2. From the trigonometric ratios of 30°, find sin 60°, cos 60°, tan 60°. 

Since 60° = 2 x 30°, from [13], [14], [15], we obtain 
sin 60° = 2 sin 30° cos'30° = V3/2. 
cos 00° = cos 2 30° - sin 2 30° = 1/2. 

tan 60°= 2ten30 ° 1 2 ^ 3 = V 3. 
1- tan 2 30° l-(V3/3) 2 

3. From the trigonometric ratios of 60°, find sin 120°, cos 120°, tan 120°. 
From [13], [14], [15], 

sin 120° = 2 sin 60° cos 60° = V3 (1/2) = V3/2. 

cos 120° = cos 2 60° - sin 2 60° = 1/4 — 3/4 = - 1/2. 

. 1ono 2 tan 60° 2 V3 /Q 

tan 120° = = — ^— = — V3. 

1 - tan 2 60° 1-3 

4. Express sin 6 A, cos 6 A, tan 6 A in terms of the trigonometric 
ratios of SA. 

From [13], [14], [15], 

sin 6 A = 2 sin 3 A cos 3 A. 

cos 6 A = cos 2 3 A - sin 2 3 A = 1 - 2 sin 2 3 A = 2 cos 2 3 J. - 1. 

tan6A = 2 tan 3 ^4/(1 - tan 2 3 A). 

5. Express sin 3 A, cos 3 A, tan 3 .A in terms of the trigonometric 
ratios of 3 A/2. 

From [13], [14], [15], 

sin 3 A = 2 sin (3 A/2) cos (3 A/2). 

cos 3 A = cos 2 (3 A/2) - sin 2 (3 A/2) = 1-2 sin 2 (3 A/2) 

= 2 cos 2 (3 A/2) -1. 
tan 3 A = 2 tan (3 A/2)/[l - tan 2 (3 A/2)]. 

~ ^ rt „ cot 2 A - 1 

6. Prove cot 2 A 



cot2A = 



2 cot A 

1 l-tan 2 A cot 2 A cot 2 A-l 



tan 2 A 2 tan A cot 2 A 2 cot A 

7. Prove esc 2 A = esc A sec A/2. 

esc 2 A = 1/sin 2 A = 1/(2 sin A cos A) = esc A sec A/2. 

8. Prove sin 2 A = (1 - cos 2 A)/2. 

From [14], cos2 A = 1 - 2 sin 2 A. 

.-. sin 2 A = (l- cos2A)/2. 



EXERCISE XIV 47 



9. Prove cos* A = (1 + cos2J.)/2. 

From [14], cos 2 A = 2 cos 2 A - 1. 

.-. cos 2 J. = (l + cos2^4)/2. 

,„ _ „ . sec 2 A 1 + tan 2 ^4 

10. Prove sec 2 A = = 

2 - sec 2 ^. l-tan 2 J. 

sec 2 ^ = 1/cos 2 A = 1/(2 cos 2 A - 1) 
sec 2 A ^1 + tan 2 ^ 
~~ 2 - sec 2 A = 1 - tan 2 ^. ' 

11. Prove cos 4 .4 = 8 sin* A - 8 sin 2 A + 1. 

cos 4-4 = 2 cos 2 2A — 1 

= 2(l-2sin 2 ^4) 2 -l 
= 8sin 4 J.-8sin 2 ^4 + 1. 

12. Prove sin 4 A = 4 sin 4 cos A — 8 sin 3 A cos A. 

sin 4 ^-4 = 2 sin 2 ^4 cos 2 A 

= 4 sin J. cos J.(l — 2 sin 2 .4) 
= 4 sin J. cos ^4 — 8 sin 3 A cos J.. 



EXERCISE XIV 

1. State in words identities [16], [17], and [18]. 

• & i* , ,1 — cos angle _.__ 

sin ftaZ/ an angle = square root of — [16] 

». ?r 7 4. - 1 + cos angrZe rirT .. 

cos half an angle = square root of — • [17] 

tan half an angle = square root of — — ■ [18] 

1 + cos angle 

2. Find sin 22i°, cos 22i°, tan 22|°, from cos 45°. 
cos 45° /l - V2/2 V2-V2 



. 0010 II- cos 45° /I 

sm22io.^__ = ^ 

COS22- = J 1 + C 2 ° s45 ° = J l+ f /2 = V¥ f^ 2 - 

t,n22r = J l - C0 ^° = J 1 -^ 2 =J 2 ^^=V^2-^2. 
2 \l + cos45° \l+V2/2 \2 + V2 



48 PLANE TRIGONOMETRY 

3. Eind sin 15°, cos 15°, tan 15°, from cos 30°. 

cos 30° /l - V3/2 V2-V3 



sin 15 c 
cos 15 c 



/l - cos 30° _ /l - -v 
\ 2 "\ 2 

/ l + cos 30° _ /l + V3/2 V2 + V3 



2 2 



1 - cos 30° 1 - V3/2 /- — — 

tan 15° = A = A / -^-V = V7 - 4 V3. 

\l + cos30° \l+V3/2 

4. cos A = 1/3 ; find the sine, cosine, and tangent of A/2. 
. A /l -cos A 11-1/3 II V3 

Sm 2 = V~^ = \^ = V3 = X' 
J. /1 + cosJ. /1 + 1/3 (2 V6 

C ° S 2 = V - ^^ = V^ = \3 = X- 

^4 _ /l - cos^4 _ 'fir- 1/3 _ /l 
tan 2" " Vl + cosJ. ~ \l + l/3 ~ \2 ~ 



V2 
2 



5. cos ^4 = a ; find the sine, cosine, and tangent of A/2. 



sin (A/2) = V(l - a)/2. 
cos (4/2) = V(l + a)/2. 
tan (4/2) = V(l - a) /(I + a). 

6. Express sin J., cos A, and tan -4. in terms of cos 2 A. 
Since A = (2 4)/2. 

.-. sin 4 = V(l- cos 2 4)/2. 
cos 4 = V(l+ cos2 4)/2. 
tan A = V(l - cos 2 2)7(1 + cos 2 4). 

7. Express sin 2 J., cos 2 A, and tan 2 4 in terms of cos 4 A. 
Since 24 = (4 4)/2. 

.% sin 2 4 = V(l - cos 4 4)/2. 
cos 2 4 = V(l + cos4 4)/2. 



tan 2 J. = V(l - cos 4 A) /(I + cos 4 A). 

8. Express sin 3 A, cos 3 A, and tan 3 A in terms of cos 6 A. 
Since 34 = (6 4)/2. 

.-. sin 3 A = V(l- cos6 4)/2. 

cos 3 4 = V(l + cos64)/2. 

tan 3 A = V(l - cos 6 A)/(l + cos 6 J.). 



EXERCISE XIV 49 

A l sinJ. \ 2 /1 + cosJA 2 

9. Prove cot 2 — = (- )=( : — — - ) • (1) 

2 \1 - cos A) \ sin A J w 

A 1 1 + cos A 

By [18] , cot 2 - = ■ - = (2) 

J L -" 2 tan 2 (^l/2) 1-cosA v ' 



1 — cos 2 A 



I sin A y 
\1 — cos A) 



(3) 
(1 -cos^) 2 U-cos^i/ v ' 

1 + cos A _ (1 + cos J.) 2 _ A + cosily 

1 - cos ^4 ~~ 1 -cos 2 A ~ \ sin ^4 / ' ( 

From (2), (3), (4), we have (1). 

10. Prove tan^ = ( anA Y= f 1 = C0S ^ Vee (csc ,4 - cot^) 2 . (1) 
2 \l + cos^./ \ sin^L / v ' v ' 

„^4 1 — cos ^4 1 — cos 2 J. / sin .A \ 2 

tan 2 — = = = • (2) 

2 1 + cos A (1 + cos A) 2 \1 + cos A) 

1 - COS ^ _ (1 - COS v4) 2 _ A - cos -A X 2 

1 + cos ^4 — 1 — cos 2 .A ~~ \ sin A / 

= (esc A -cot A) 2 . (3) 

From (2) and (3) we obtain (1). 



-,-, -d » A 2sec^L 

11. Prove sec 2 — = 

2 sec A + 1 


sec A _ 2 sec J. 


2 cos 2 (^4/2) 1 + cos^4 
-. « ^, » ^ 2 sec ^4 


sec ^4 ~~ sec A + 1 


2 sec A — \ 

2 A - 1 2 


sec J. 2 sec -4 


2 sin 2 M /2) 1 - cos J. 


sec A sec A — 1 



13. Express cos 4 J. in terms of cos 2 J. and cos 4 ^4. 

(cos 2 J.) 2 ee (i + i cos 2 ^4) 2 

= ^ + icos2^4 + | cos 2 2 ^4 

ee | + I cos 2 4 + I (I + i cos 4 A). 

.-. cos 4 J. ee | + I cos 2 A + i cos 4 A. 

14. Prove sin 4 J. ee | — | cos 2A + \ cos 4 .4. 

(sin 2 AY ee (i - i cos 2 ^4) 2 

ee i " — i cos 2 A + \ cos 2 2 A 

= I - | cos 2 J. + i (J + i cos 4 A) 

= 4 - i cos 2 J. + | cos 4 A. 



50 PLANE TRIGONOMETRY 

15. Prove sin 2 A cos 2 A = i — | cos 4 ^1. 

sin 2 Jl cos 2 J. = (2 sin A cos ^4) 2 /4 

= £sin 2 2J.= i -i cos 4^1. by [16] 

16. Prove sin 2 A cos 4 A = ^ + | sin 2 2 J. • cos 2 ^i - T ^ cos 4 J.. 

sin 2 A cos 4 ^4. = \ (2 sin J. cos A) 2 cos 2 J. 

= | (sin 2 2 ^L) [(1 + cos 2 J.)/2] 

= | (sin 2 2 ^. + cos 2 ^4 sin 2 2 J.) 

= i [(1 - cos 4 A)/2 + cos 2 J. sin 2 2 A] 

— Tff + 1 sin 2 2 ^4 cos 2^4. - T ^ cos 4 J.. 

EXERCISE XV 

1. State in words identities [19] ■ • ■ [22]. 

twice sin half sum into cos half difference. 



The sum of the sines of any "1 



two angles * j 

The difference of the sines 1 . - . , _ 

„ , , ^ = twice cos /icu/ sum into sin half difference, 

of any two angles J 

The sum of the cosines ofl ,- mj-« 

}- = twice cos half sum into cos half difference, 
any two angles j 

The difference of the cosines^ . ' . ,<,,.«. 

, , , y = — twice sm/iaZ/ sum into sm/ku/difference. 

of any iwo angles J 

2. Prove sin 60° + sin 30° = 2 sin 45° cos 15°. (1) 
Let C = 60°, D=30°; 

then (C + D)/2 = 45°, 

(C - D)/2 = 15°. 
Substituting these values in [19], we obtain (1). 

3. Prove sin 50° + sin 10° = 2 sin 30° cos 20°. (1) 
Let C=50°, D = 10°; 

then (C + D)/2 = 30°, 

(C - D)/2 = 20°. 
Substituting these values in [19], we obtain (1). 

4. Prove cos 75° + cos 15° = 2 cos 45° cos 30°. (1) 
Let = 75°, D = 15°; 

then (C + D)/2 = 45°, 

(C - D)/2 = 30°. 
Substituting these values in [21], we obtain (1). 



EXERCISE XV 51 

5. Prove cos 80° - cos 20° = - 2 sin 50° sin 30°. (1) 

Let C = 80°, D = 20°; 

then (G + D)/2 = 50°, 

(C - D)/2 = 30°. 

Substituting these values in [22], we obtain (1). 

6. Prove sin 7 A — sin SA = 2 cos 5 A sin 2 A. (1) 

Let C = 7.4, J> = 3^.; 

then (C + D)/2 = 5A, 

(C-D)/2 = 2A. 

Substituting these values in [20], we obtain (1). 

„ _ sin 1 A — sin5J. . 

7. Prove = tan .4. 

cos 1A+ cos 5^4 

Let C = 7A, D = 5A;. 

then (C + D)/2 = 6A, 

(C - D)/2 = A. 

By [20] and [21] we obtain 

sin 7 A — sin 5 A _ 2 cos 6 J. sin A 
cos 7 J. + cos 6 A 2 cos 6 A cos .4 
= tan A. 

_ „ sin ^4. + sin 3 J. _ . 

8. Prove = tan 2 ^4. 

cos A + cos 3 A 
Let C = 3J., D = ^4; 

then (C + D)/2 = 2^L, 

(C-D)/2=A. 

By [19] and [21] we obtain 

sin 3 A + sin A _ 2 sin 2 J. cos A 
cos 3 A + cos A 2 cos 2 A cos J. 
= tan 2 A. 

„ n sinC + sinD tani(C + D) 

9. Prove = — -• 

sin C — sin D tan \ (C — D) 

sin C + sin D = 2 sin i (C + D) cos i (C - D) 
sin C - sin D _ 2 cos i (C + D) sin i (C - D) 

= tani(C + D) 

~~ tan|(C- D)' 



52 PLANE TRIGONOMETRY 



, ^ ■ sin C + sin B C + B 

10. Prove = tan — 

cosC+cosD 2 

By [19] and [21], **<* + «»» ee 8"" i(C + J)o«HC- Z» 

JL J L J cosC + cosD 2 cos £ (C + Z>) cos i(C-Z>) 
= tan i (C -f D). 

„ „ sinC + sinD C-B B-C 

11. Prove = — cot = cot 



cosC-cosD 2 2 

By [19] and [22], si ^±_ si ^ ^ 2 sin H C + B) cos H C - B) 
J L J L J 'cosC-cosD -2sini(C + D)sini(C-D) 
= - cot !(<?-£>) 
= coti(D- C). 

-, „ t, sin C - sin D C - D 
12. Prove ■ = tan 

cos C + cos B 2 

sin C - sin D _ 2 cos \ r (C + D) sin i(C - Z>> 



By [20] and [21] 



cos C + cosD 2 cos | (C + -D) cos i (C - B) 
= tan i(C-D). 



• _ sinC-smD ,C + B 

13. Prove = - cot 

cos C — cos D 2 

By [20] and [22], ^-sinD s 2 cos H C + i» sin i(C - D) 
J L J L J 'cosC-cosD - 2 sin i(C + D) sin i(C-D) 
= -cot\{C-\-B). 

, *. „ cos C + cos B C + B B-C 

14. Prove = cot cot 



By [21] and [22], 



cos C — cos B 2 2 

cos C + cos D _ 2 cos |(C + D) cos \ {C - B) 



cos C - cos D — - 2 sin i ( C+ D) sin \(C - B) 
= - cot | (C + B) cot i (C - B) 
= cot HC + D) cot i(i>- C)- 



15. Given sin J. = 1/2, sin B = 1/3, to find sin (.4 + B), sin {A - B), 
cos (A + 5), cos (J. - .B), sin 2 J., sin 2 B, cos 2 A, coa 2 B : (1) when J. 
and B are both in the first quadrant ; (2) when A is in the first and B is 
in the second quadrant. 

(1) When A and B are both in the first quadrant. 

cos A = Vl - 1/4 = V3/2. 
cos B = Vl - 1/9 = 2 V2/3. 



EXERCISE XV 53 

Substituting the values of sin A, cos A, sin B, cos B in [7], [8], [11], [12], 

ri51, [16], we obtain 

sin {A + B) = (2 V2 + V3)/6, 
sin (J. - 5) = (2 V2 - V3)/6, 
cos (.4 + B) = (2 V6 - l)/6, 
cos (A - B) = (2 V6 + l)/6, 

sin2^4=V3/2, 

sin2J5=4V2/9, 

cos2^i = l/2, 

008 21? = 7/9. 

(2) When J. is in the first and B is in the second quadrant, 
cos A = Vl -1/4 = V3/2. 
cos 5 = - Vl _ 1/9 = - 2 V2/3. 

Substituting the given values of sin A and sin B, and these values of 
cos J. and cos B in [7], [8], [11], [12], [15], [16], we obtain 
sin (A + B) = - (2 ^2 - V3)/6, 
sin (4 - B) = - (2 V2 + V3)/6, 
cos (J. + B) = - (2 V6 + l)/6, 
cos (J. - B) = - (2 V6 - l)/6, 

sin 2 A = V3/2, 

cos2^4 = l/2, 

sin 2 5 = - 4 V2/9, 

cos 2 5 =7/9. 

16. From the answers to example 15 find in the simplest way 
tan (A + B), tan (A - B), cot (A + B), cot (A - B), sec (A + B), 
esc (J. + B), tan 2 J., cot 2 J., sec 2 5, esc 2 5, in cases (1) and (2). 

(1) When A and B are both in the first quadrant. 

tan (A + B) = ™(^ + *> = 2V2 + ^ 3 . 

cos (A + 5) 2 V6 - 1 
tan (A - 5) = (2 V2 - V3)/(2 V^ + 1). 
cot (A + B) = 1/tan (A + 5) 

= (2V6-1)/(2V2+V3). 
cot (A-B) = (2 V6 + l)/(2 V2 - V3). 
sec (.4 + B) = 1/cos (J. + B) 
= 6/(2V6-l). 
esc (4 + B) = 6/(2 V2 + V3). 

tan 2 A = sin 2 A/cos 2A = s/Z. 
cot 2 J. =1/V3. 
860 2 5=1/008 2 5 = 9/7. 
esc 2B = 9/(4 V2). 



54 PLANE TRIGONOMETRY 

(2) When A is in the first and B is in the second quadrant. 

v ' cos {A + B) 2 V<3 + 1 

tan (J. - 5) = (2 V2 + V3)/(2 V6 - 1). 
cot (A + B) = 1/tan (J. + 5) 

= (2V6+1)/(2V2-V3). 
cot (A - B) = (2 V6 - l)/(2 V2 4- V3). 
sec (J. + B) = 1/cos (J. + S) = - 6/(2 V6 + 1). 
esc (^ 4- B) = - 6/(2 V2 - V3). 

tan 2 ^4 = sin 2 .4 /cos 2A = V3. 

cot 2^1 = 1/V3. 

sec 2B = 9/7. 

esc 2 5 = - 9/(4 V2). 



EXERCISE XVI 

, ^ sin (x + y) tan x + tan v 

1. Prove — = • 

sin (x — y) tan x — tan y 

sin (x + y) _ sin x cos y + cos x sin y 
sin (x — y) ~ sin x cos y — cos x sin y 
sin x sin y 
_ cos x cos y _ tan x + tan ?/ 
— sin x sin ?/ — tan x — tan y 
cos x cos y 

„ ^ cos (x + ?/) 1 — tan x tan y 

2. Prove — = 

cos (x — ?/) 1 + tan x tan y 

cos (x + ?/) _ cos x cos y — sin x sin y 
cos (x — y) ~~ cos x cos y 4- sin x sin ?/ 

sin x sin y 

cosx cosy _ 1 — tan x tan y 
~ sin x sin y ~ 1 + tan x tan ?/ 

cosx cosy 

_ _ cos (x + 2/) 

3. Prove — = cot x — tan ?/. 

sin x cos ?/ 

cos (x 4- y) _ cos x cos y — sin x sin y 
sin x cos y sin x cos ?/ 

= cot x — tan y. 



EXERCISE XVI 55 



COS (X — v) 

4. Prove — = tan x + cot y. 

cos x sin y 

cos {x. — y) _ cos x cos y + sin £ sin ?/ 





cos x sin ?/ cos x sin y 




= cot ?/ + tan x. 


5. 


_ , 2 -sec 2 J. 

Prove cos 2 A = • 

sec 2 .4 


Bi 


j [14], cos 2 A = 2 cos 2 ^4-1 




- 2 i 



sec 2 ^4 

_2- sec 2 ^4 
■ — sec 2 ^4 

^ -o « „ csc 2 ^4 

6. Prove sec 2 ^4 = 

csc 2 ^4 - 2 

sec 2 J. = 1/cos 2 .4 

= 1/(1 -2 sin 2 ^4) 
= 1/(1 - 2/csc 2 ^4) 
csc 2 ^4 

_ CSC 2 ^i -2* 

r, -r, • « >« 2tan^. 

7. Prove sin. 2 J. = 

1 + tan 2 ^4 

sin 2 A = 2 sin J. cos ^4 
= 2 tan J. cos 2 J. 
= 2 tan J. /sec 2 J. 
= 2 tan .4/(1 + tan 2 ^4). 

_ _, sin 3 A — sin A . 

8. Prove = tan A. (1) 

cos3J. + cosJ- 

If C = 3Aan&D = A, then (C + D)/2 = 2^4, (C - D)/2 = J.; hence, 
by [20] and [21], we obtain 

sin 3 A — sin A 2 cos 2 ^4 sin J. 



cos 3 A + cos J. 2 cos 2 ^4 cos A 



tan J.. 



9. Express sin(3x/4), cos(3x/4), and tan(3x/4) in terms of cos(3x/2). 

-o ri«n . 3x /l- cos (3 a/2) 

By [16], sin 



. 3 x / 1 

in — = ^ /— 

4 \ 



By [17], cos ^ = ^- 

3x_ /l-cos(3x/2) 



2 
3x_ + cos (3 z/2) 

2~ 



By [18], tan 



vs 



4 \ 1 + cos (3 x/2) 



56 PLANE TRIGONOMETRY 

10. Express sin (3 x/4), cos (3 x/4), and tan (3 x/4) in terms of the 
trigonometric ratios of 3 x/8. 

By [13] , sin (3 x/4) = 2 sin (3 x/8) cos (3 x/S) . 

By [14], cos (3 x/4) = cos 2 (3 x/8) - sin 2 (3 x/8). 

By [15], tan (3 x/4) = 2tan(3x/8)/[l - tan 2 (3 x/8)]. 

11. Prove (sin A + cos A) 2 = 1 + sin 2 A. 

(sin A + cos A) 2 = sin 2 A + cos 2 A + 2 sin A cos A 
= 1 + sin 2 J.. 

12. Prove (sin J. — cos A) 2 = 1 — sin 2 A. 

(sin J. — cos A) 2 = sin 2 J. + cos 2 J. — 2 sin A cos J. 
= l-sin24. 

13. Prove tan A + cot A = 2 esc 2 J.. 

tan ^L + cot A = (1 + cot 2 A) tan J. 
= esc 2 A tan J. 
= l/(sin^l cosvl) 
= 2/sin 2 ^4 = 2 esc 2 ^4. 

14. Prove cot A — tan A = 2 cot 2 J.. 

cot -4 — tan A = (cot 2 ^4 — l)/cot A 

= 2[(cot 2 ^4-l)/(2cot^4)] 
= 2 cot 2 ^4. 

_ _ _ tan A + tan 5 

15. Prove = tan ^4 tan 5. 

cot ^4 + cot J5 

tan J. + tan B _ tan B (tan ^4 cot 5 + 1) 
cotJ. + cot5 cotJ.(l + cot J5tan^4) 

= tan B/cot A ~ tan A tan B. 

16. Given sin A =2/3, cos B = 1/2; to find (1) sin (^4+ B), sin (A- B), 
cos (.4 + 5),cos(J.-5),sin2^4, cos 2 A, 8^25,00825; (2) tan(^4 + 5), 
cot(^4 + 5), tan(A-5), cot (A - 5), tan 2 J., cot 2 -4, tan25, cot25. 

cos A = V l — sm^ A 

= Vi Z 4/9 = V5/3. 
sin 5 = Vl - cos 2 B 

= Vl - 1/4 = V3/2. 
tan ^4 = sin J./cos^4 = 2/ V&- 
cot ^4 = 1/tan A = V5/2. 
tan B = sin 5/cos B = V3- 
cot5 = l/tan5 = l/V3. 



EXERCISE XVI 57 

(1) Substituting the values of sin J., cos ^4, sin B, cos B in [7], [8], [9]. 
[10], [13], [14], we obtain 

sm(^ + £) = (2+V15)/6, 
sin(^-5) = (2-V15)/6, 
cos(^ + B) = (V5 - 2 V3)/6, 
cos (J. - B) = (Vo + 2 V3)/6, 

sin2^4 = 4V5/9, 

cos 2 ^4 = 1/9, 

sin 2 5 = V3/2, 

cos 2 5 = -1/2. 

(2) Substituting the values of tan A, cot A, tanJ5, cot B in [11], [12], 
[15], and in the formulas in examples 8 and 9 in Exercise XII, we obtain 

2/V5+V3 2+V15 



tan (A + B) = 
cot(A + B) = 



l-(2/V5)V3 V5-2V3 
1 V5 - 2 V3 



tan (A + B) 2 + V15 
tan (A -B) = {2- y/lb)/(y/5 + 2 V3), 
cot (.4 -B) = (V5 + 2 V3)/(2 - V15), 

tan 2 ^4 = 4 V5, 

cot 2 ^4 = V5/20, 

tan2£ = - V3, 

cot 2 B = - V3/3. 

17. Prove cot ^ + cot£ = csc {B _ ^ sin (5 + A y 
cot A- cot B v /at; 



By [3], 



cot A + cot B _ cos ^4 sin i? + cos 5 sin J. 

cot J. — cot B cos ^4 sin B — sin J. cos B 
_ sin (£ + A) 
~ sin (J5 - A) 
= csc (5 - A) sin (5 + A). 

, „ „ sin ( J. + 5) sin ( J. - #) 

18. Prove — * — ^— -; J * = tan 2 J. - tan 2 B. 

cos 2 J. cos 2 J5 

sin (A + j) sin (J. - B) 

cos 2 J. cos 2 B 

_ (sin A cos -B + cos A sin J5) (sin A cos 5 — cos A sin J5) 

cos 2 A cos 2 5 

sin 2 J. cos 2 .B — cos 2 A sin 2 5 



cos 2 ^4 cos 2 B 



= tan 2 ^4 -tan 2 B. 



58 PLANE TRIGONOMETRY 



19. Prove tan ^ = tan ^ = tan (A + B) tan (A - B). 
l-tan 2 ^Ltan 2 £ . 

tan 2 A — tan 2 B tan A + tan 5 tan A — tan jB 

x 



1 - tan 2 A tan 2 B 1 - tan ^4. tan B 1 + tan J. tan B 
= tan (^i+ B) tan (/!-£). 

20. Prove V2 sin (J. ± 45°) = sin ^4 ± cos J.. 

V2 sin (^4 ± 45°) = V2 (sin ^4 cos 45° ± cos ^4 sin 45°) 
= sin A ± cos A. 

21. Prove 2 sin (45° - A) cos (45° + B) = cos (A - B) - sin (J. + J5). 

2 sin (45° - ^4) cos (45° + B) = 2 (sin 45° cos J. - cos 45° sin ^4) 

(cos 45° cos B — sin 45° sin B). 
= (cos A — sin ^4) (cos B — sin B) 
= cos A cos J5 + sin ^4 sin I? 

— sin A cos 2? — cos A sin i? 
= cos (A- B)- sin (^4 + -B). 

22. Prove 2 sin (45° + A) cos (45° + B) = cos (^4 + B) + sin (J. - B). 

2 sin (45° + ^4) cos (45° -f JB) = 2 (sin 45° cos A + cos 45° sin ^4) 

(cos 45° cos B — sin 45° sin B) 
= (cos A + sin A) (cos B — sin B) 
~ cos J. cos B — sin A sin J5 

-f sin A cos jB — cos A sin 2? 
= cos (J. -f B) + sin (J. - B). 

23. Prove 2 sin (45° + A) cos (45° - B) = cos {A - B) + sin (^4 + £). 

2 sin (45° + ^4) cos (45° - B) = 2 (sin 45° cos A + cos 45° sin A) 

(cos 45° cos B + sin 45° sin B) 
= (cos J. + sin ^4) (cos B + sin B) 
= cos ^4 cos JB + sin A sin 2? 

-f sin A cos B + cos ^4 sin J5 
eecos (A - B) + sin (J. + 5). 



24. Prove cot (A + 45°) = ^^ — - = \- 
K ' cot^ + 1 \1 

cot (A + 45°; 



EXERCISE XVI 59 

sin 2 A 1 — sin 2 A 



cot A + 1 \ 1 + sin 2 J. cos 2 J. 
cot A cot 45°- 1 



cot A + cot 45° 
cot A — 1 



cot A + l 

cos J. — sin A 

cos J. + sin A 
/cos 2 J- + sin 2 J. — 2 sin J. cos A 
cos 2 J. + sin 2 A + 2 sin J. cos A 
1 - sin 2 A 



■V 
■V 



1 + sin 2 A 
(l-sin2^1) 2 

1 - sin 2 2 J 

sin 2 J. 



cos 2 A 



«r -r. x/^ ,4 cm cot .4 + 1 tan .4 + 1 

25. Prove cot (A - 45°) = — = — • 

1 - cot A tan A - 1 

, rm cot A cot 45° + 1 

cot (A - 45°) = ■ — 

cot 45° - cot A 

_ cot A + 1 tan A 

~ 1 — cot A tan A 

1 + tan A 



tan A — 1 



26. Prove tan (J ± 45°) + cot (A + 45°) = 0. 

, rnN x/ . , rnx tanJ. + l cotJ.il tan J. 

tan (A ± 45°) + cot (J. + 45°) = — + 

1 + tan J. 1 =f cot A tan A 

_ tan A ±1 1 ± tan J. 

— 1 + tan A tan J. + 1 

_ tan 4±lTl- tan J. _ 

1 + tan J. 

27. Prove sin 9 £ — sin 7x = 2 cos 8 x sin x. 

-d rom • n ■ r, o 9X + 7X.9£-7X 

By [20], sin 9 x — sin 7 z = 2 cos sin 

= 2 cos 8 x sin cc. 



60 PLANE TRIGONOMETRY 



28. Prove cos 7 x + cos 5 x = 2 cos 6 x cos x. 

-orom if-'", r o 7x+5x 7x-5x 

By [21] , cos 7 x + cos 5 x = 2 cos cos 

= 2 cos 6 x cos x. 

« rt ^ sin3x-sinx 

29. Prove = cot 2 x. 

cosx — cos3x 



By [20] and [22], 

sin 3 x — sin x 2 cos 2 x sin x 



= cot 2 x. 



cos x — cos 3 x 2 sin 2 x sin x 

„ sin 5 x — sin 2 x 7x 

30. Prove = cot 

cos 2 x— cos 5 x 2 

By [20] and [22], 

sin 5 x — sin 2 x _ 2 cos (7 x/2) sin (3 x/2) 

cos 2 x — cos 5 x — 2 sin (7 x/2) sin (3 x/2) 

= cot (7 x/2). 

sin A + sin B cos A + cos B 



31. Prove 



cos A — cos B sin B — sin A 



sin A + sinB cos J. + cos B _ sin 2 1? + cos J5 2 — (sin 2 A + cos 2 J.) _ 
cos^l — cos 5 sin B — sin J. — (cos A — cos 5) (sin B — sin J.) 

32. Prove tan (x/2 +.45°) = tan x + sec x . 

, - , rnx tan (x/2) + tan 45° 

tan (x/2 + 45° = v ; 

v ; 1 - tan (x/2) tan 45° 

_ 1 + tan (x/2) 

~" 1 - tan (x/2) 

_ cos (x/2) + sin (x/2) 

— cos (x/2) — sin (x/2) 

_ cos 2 (x/2) + sin 2 (x/2) + 2 sin (x/2) cos (x/2) 

= cos 2 (x/2) - sin 2 (x/2) 

_ 1 + sin x 

cosx 
= secx + tanx. 

33. Find sin A and cos A when they are proportional to any two real 
numbers, as a and b. 

Let sin A = a/ Va' 2 + & 2 . 

Then, by § 24, cos A = b/Va* + 6 2 . 

Hence sin A/a = cos A/b. 

.-. sin A :cos^l = a : 6. 



EXERCISE XVII 



61 



34. Find sin A and cos J. when they are proportional to 3 and 4. 

Here a = 3, b = 4, Va 2 + 6 2 = V3 2 + 4 2 = 5. 
Hence sin A = 3/5 and cos .A = 4/5. 



35. If 



cos 2 A + cos 2 B + cos 2 = 1, 



(1) 



find the value of cos A, cos 2?, and cos C when they are proportional to 
any three real numbers, as a, 6, and c. 

Let cos J. = a/^/a? -f b 2 + c 2 , 

and cos B = b/ Va 2 + 6 2 + c 2 . 

cos C = c/Va 2 + & 2 + c 2 . 



Then by (1), 
Hence 



cos A/a = cos jB/6 = cos C/c. 
.-. cos ^L : cos B : cos C = a : b : c. 



EXERCISE XVII 

Solve each of the following triangles in which C = 90° : 
1. Given B = 67°, a = 5. 

A = 90° - B = 23°. 

b = a tan 5. c = a/cos 5 

loga = 0.69897 loga = 10 

log tan B = 0.37215 log cos B = _9 

.-. log 6 



1.07112 
.-. b= 11.779. 

2. Given J. = 38°, a = 8.09. 

5 = 90° - A = 52°. 
b = a/t&nA. 
loga = 10.90795 -10 
log tan A = 9.89281 - 10 
.-. log b= 1.01514 
.-. b = 10.355. 

3. Given ^ = 15°, c = 7. 

B = 90° - A = 75°. 

a = csin^i. 

logc = 0.84510 

log sin A = 9.41300 - 10 

.-. log a = 0.25810 

.-. a= 1.8117. 



.59188 
logc= 1.10709 



•. c = 12.796. 



c = a/sin A. 
loga = 10.90795 -10 
log sin A = 9.78934 -10 
.-. logc= 1.11861 
.-. c = 13.14. 



b = c cos A. 
logc = 0.84510 
log cos A - 9.98494 - 10 
.-. log 6 = 0.83004 
.-. b = 6.7614. 



62 



PLANE TRIGONOMETRY 



4. Given B = 50°, b = 20. 






A = 90° - B = 40°. 






a = b cot J5. 


c = &/sin B. 




log 6 = 1.30103 


log 6 = 11.30103 • 


-10 


log cot I? = 9.92381 -10 


log sin B= 9.88425- 


-10 


.-. loga = 1.22484 


.-. logc = 1.41678 




.-. a = 16.782. 


.-. c = 26.108. 




5. Given a = 0.35, c = 0.62. 






sin A = a/c. 






loga = 19.54407 -20 


b = c sin 5. 




logc = 9.79239-10 


log c = 9. 79239 - 


10 


.-. log sin A= 9.75168-10 


log sin B = 9.91668 - 


-10 


.-. A = 34° 22' 9". 


.-. log b= 9.70907- 


10 


£ = 90° - A = 55° 37' 51". 


.-. b = 0.51176. 




6. Given a = 273, 6 = 418. 






tan A = a/b. 






loga = 12.43616 -10 


c = a/ sin ^.. 




log b= 2.62118 


log a = 12.43616 


-10 


.-. log tan A- 9.81498-10 


log sin A= 9.73784 


-10 


.-. A = 33° 8' 56". 


.-. logc= 2.69832 




B = 90° - A = 56° 51' 4". 


.-. c = 499.26. 




7. Given 6 = 58.6, c = 76.3. 






sin B = 6/c. 






log 6 = 11.76790 -10 


a = csinA. 




logc= 1.88252 


logc = 1.88252 




.\ log sin B= 9.88538-10 


log sin ^1 = 9.80646 - 


10 


.-. B = 50° 10' 38". 


.-. loga = 1.68898 




A = 90° - B = 39° 49' 22". 


.-. a = 48.863. 




8. Given A = 9°, b = 937. 






£ = 90° - J. = 81°. 






a = b tan -4. 


c = 6/cos-4. 




log 6 = 2.97174 


log 6 = 12.97174 ■ 


-10 


log tan ^1 = 9.19971 -10 


log cos A = 9.99462- 


-10 


.-. loga = 2.17145 


.-. logc= 2.97712 




.-. a = 148.41. 


.-. c = 948.68. 





EXERCISE XVII 



63 



9. Given a = 3.414, b = 2.875. 
tan A = a/b. 
\oga = 0.53326 
log b = 0.45864 
.-. log tan ^1 = 0.07462 

.-. A = 49° 53' 53". 
B = 90° - A = 40° 6' 7". 

10. Given A = 46° 23', c = 5278. 
B = 90° - A = 43° 37'. 
a = csin.4. 
logc = 3.72252 
log sin A = 9.85972 - 10 
.-. loga = 3.58224 
.-. a = 3821.5. 



c = a/ sin J.. 
loga = 10.53326 -10 
log sin A = 9.88361 -10 
.-. logc= 0.64965 
.-. c = 4.4632. 



6 = ccos A. 

logc = 3.72252 

log cos A = 9.83874 - 10 

.-. log 6 = 3.56126 

.-. b = 3641.3. 



11. Given a = 529.3, c = 902.7. 

sin A = a/c. 

loga = 12.72370 -10 

log c = 2.95554 
.-. log sin A = 9.76816-10 

.-. A = 35° 53' 56.5". 
5 = 90° - A = 54° 6' 3.5". 



6 = ccos A. 

logc = 2.95554 

log cos ^L = 9.90852 - 10 

.-. log& = 2.86406 

.-. b = 731.23. 



12. Given 5 = 23° 9', b = 75.48. 
A = 90° - B = 66° 51'. 
a = b cot 5. 
log 6 = 1.87783 

log cot B = 0.36899 

.-. loga = 2.24682 

.-. a = 176.53. 



c = 6/sin B. 

log& = 11.87783 -10 
log sin B = 9.59455 - 10 
\ logc= 2.28328 
.-. c = 191.99. 



13. Given B = 18° 38', 


c = 2.5432. 






A = 90° - B 


= 71° 22'. 






b = c sin 5. 




a = c cos B. 




logc = 0.40538 




logc = 0.40538 




log sin B = 9.50449 - 


-10 


log cos 5= 9.97662 - 


-10 


.-. log 6 = 9.90987- 


-10 


.-. loga = 0.38200 




.-. b = .81258. 




.-. a = 2.4099. 





64 



PLANE TRIGONOMETKY 



14. Given A = 31° 45', a = 48.04. 

B = 90° - A = 58° 15'. 
b = a cot A. 
loga = 1.68160 
log cot A- 0.20844 
.-. log 6= 1.89004 
.-. b = 77.632. 

15. Given b = 617.57, c = 729.59. 

sin B = b/c. 
log 6 = 12.79069-10 
log c = 2.86308 
.-. log sin B= 9.92761 -10 
.-. B = 57° 49' 45 // - 
A = 90° - B = 32° 10' 15 77 . 



c = a /sin. A. 
log a = 11.68160 -10 
log sin A = 9.72116 -10 
.-. logc= 1.96044 
.-. c = 91.294. 



a = c cos B. 

logc = 2.86308 
log cos B = 9.72627-10 
.-. loga = 2.58935 
.-. a = 388.46. 



16. Given B = 82° 6' 18' 

A = 90° - B = 7° 53' 42". 

6 = a tan 5. 

loga= 1.95095 

log tan B = 0.85801 

.-. log& = 2.80896 

.-. 6 = 644.11. 



32. 



c = a/cos .B. 

loga = 11.95095 -10 

log cos B = 9.13786-10 
.-. logc = 2.81309 
.-. c = 650.27. 



EXERCISE XVIII 

We shall use the following notation in isosceles triangles 

r = one of the equal sides, 
c = base, 
h = altitude, 

A = one of the equal angles, 
C = angle at the vertex, 
Q = area of the triangle. 

1. Given c and A, find C, r, h. 

C = 180°-2J.. 

r — c sec A /2 = c/(2 cos vl). 
h = ctan.A/2. 




EXERCISE XVIII 65 

2. Given ft and C, find A, r, c. 

4 = 90° - C/2. 
r = A sec (C/2) = A/cos (C/2). 
c = 2ft tan (C/2). 

3. Given c and ft, find .4, C, r. 

tan A = 2 ft/c, or A = tan- x (2 h/c). 
C = 2tan~i(c/2/i). 
r = ft esc J. = ft /sin ^ , 
or r = Vft2 + c 2 /4 = V4 ft 2 + c 2 /2. 

4. Given c = 2.352, C = 69° 49', find r, ft, -4, Q. 

^ = (180° - C)/2 = (180° - 69° 490/2 = 55° W 30". 
2 Q = eft. 

2 r = c sec J. = c/cos -4. 2 ft = c tan ^4. 

log c = 10.37144 - 10 log c = .37144 

log cos A = 9.75760 -10 log tan A = .15626 

.-. log2r= .61384 .-. log2ft = .52770 

.-. r = 4.11/2 = 2.055. .-. ft = 3.3706/2 = 1.6853. 

logc = .37144 
log ft = .22667 
.-. log2Q = .59811 

.-. Q = 3.9638/2 =1.9819. 
'5. Given ft = 7.4847, A = 76° 14', find r, c, C, Q. 
C = 180° - 2 ^4 = 27° 32'. 
Q = ftc/2, 
r = ft/sin .A. c/2 = ft cot A. 

log ft = 10.87417 - 10 log ft = 10.87417 - 10 

log sin A = 9.98734-10 log cot A = 9.38918-10 

.-. logr= .88683 .-.log (c/2) = .26335 

.-. r = 7.706. .-. c = 2 x 1.8338 = 3.6676. 

log ft = .87417 

log (c/2) = .26335 

.-. logQ = 1.13752 

.-. Q = 13.7253. 

6. A barn is 40 x 80 ft. , the pitch of the roof is 45° ; find the length 
of the rafters and the area of both sides of the roof, the horizontal 
projection of the cornice being 1 ft. 

Let x = length of rafter and Q = area of roof ; then, since the roof 
projects 1 ft. over the side of the barn, we have : 



66 PLANE TRIGONOMETRY 



x = 21/cos45°. 


Q = 2 • x • 82. 


log 21 = 11.32222 -10 


logx = 1.47273 


; cos 45°= 9.84949-10 


log 164 = 2.21484 


.-. logx= 1.47273 


.-. logQ = 3.68757 


.-. x = 29.698. 


.-. Q = 4870.44. 



7. One side of a regular decagon is 1 ; find r, ft, F. 

C/2 = 180°/n = 18°, c = 1. 2F = ph. 

2r = c/sin(C/2), 2ft = c cot (C/2). 

log c = 0. 00000 log c = . 00000 

log sin (C/2) = 9.48998 -10 log cot (C/2) = .48822 

.-. log2r= .51002 .-. log (2ft) = .48822 

.-. r = 3.23608/2 = 1.61804. .-. ft = 3.07764/2 = 1.53882. 

logp = 1.00000 
log ft = .18719 
* .-. log(2F) = 1.18719 

.-. F= 15.3882/2 = 7.6941. 

8. The perimeter of a regular dodecagon is 70 ; find r, ft, F. 

c = 70/12 = 5.833i. C/2 = 180°/n = 15°. 4F=2ph. 
2r = c/sin(C/2). 2 ft = c cot(C/2). 

logc= .76592 logc= .76592 

log sin ( C/2) = 9.41300 - 10 log cot ( C/2) = 0.57195 

.-. log2r = 1.35292 .-. log (2 ft) = 1.33787 

.-. r = 22.5384/2 = 11.2692. .-. ft = 21.7705/2 = 10.8852. 

logp = 1.84510 
log(2 7Q = 1.33787 



.-. log(4i^) = 


= 3.18297 






F = 


: 1523.96/4 = 380.99. 


9. In a regular octagon ft = 


= 1; 


find 


r, c, F. 


C/2 = 18078 = 22° 


30'. 




F=4ch. 


r = ft/cos (C/2). 






c/2 = ft tan (C/2). 


log ft = 0.0000 






log ft = 0.00000 


log cos ( C/2) = 9.96562 -10 






log tan (C/2) = 9.61722 - 10. 


.-. logr= .03438 






.-. log (c/2) = 9.61722 -10 


.-. r = 1.0824. 






.-. c = .4142 x 2 = .8284 






F = 


: 4 x .8284 = 3.3136. 



EXERCISE XVIII 



67 



10. The area of a regular heptagon is 7 ; find r, A, p. 

C/2 = 180^7 = 25° 42f. Q = 1. /* = (c/2) cot (C/2). 

... 1 = Q = he/ 2 = (c/2)2 cot (C/2), or c/2 = Vtan (C/2). 
.-. p = U (c/2) = 14 Vtan (C/2). 
log Vtan (C/2) = (9.68266 - 10)/2 r = ( c / 2 )/ sin (C/2). 

= 9.84133-10 log (c/2) = 9.84133 -10 

log 14 = 1.14613 log sin ( C/2) = 9.63737 - 10 

.-. logp= .98746 .-. logr= .20396 

.-. p = 9.7154. .-. r = 1.5994. 

log (c/2) = 9.84133 -10 

log cot ( C/2) = .31733 
.-. logh= .15866 

.-. h = 1.441. 

11. The side of a regular octagon is 24 ft. , find h and r ; also find the 
difference between the areas of the octagon and the inscribed circle, and 
the difference between the areas of the octagon and the circumscribed 
circle. 



r = (c/2)/sin (C/2). 


h = (c/2) cot (C/2). 


log (c/2) = 11.07918 -10 


log (c/2) = 1.07918 


ogsin(C/2)= 9.58284 -10 


log cot ( C/2) = .38278 


.-. logr= 1.49634 


.-. log^ = 1.46196 


.-. r = 31.357. 


.-. h= 28.971. 


Area inscribed circle = s = 7th 2 . 




Area circumscribed circle = s' = 


7tr 2 . F = ich. 


log (A 2 ) = 2.92392 


log(4c) = 1.98227 


log?r= .49715 


logh= 1.46196 


logs = 3.42107 


.-. log F= 3.44423 


s = 2636.75. 


.-. F= 2781.2. 


log(r2) 


= 2.99268 


logTT 


= .49715 


.-. logs' 


= 3.48983 


.-. s' 


= 3089.1. 


Hence s' — F= 307.9, and F - i 


s = 144.45. 



68 



PLANE TRIGONOMETRY 



12. The side of a regular heptagon is 14 ft 
in example 11. 

Here c = 14, C/2 = 180°/ 7 = 25° 42' 51". 

r = (c/2)/sin (C/2). 

log(c/2) = 10.84510 -10 
log sin (C/2) = 9.63737 -10 
.-. logr 
.-. r 



find the magnitudes as 



1.20773 
16.134. 
V = 7tr 2 . 



ft = (c/2) cot (C/2). 

log (c/2) = 0.84510 

cot (C/2) = 0.31734 

.-. log ft = 1.16244 

.-. ft = 14.536. 



F = 49 ft. 



= 7th*. 



log(r 2 ) = 2.41546 

log ?r = 0.49715 

.-. logs' = 2.91261 

.-. s' = 817.73. 



log (ft 2 ) 

lOgTT 

logs 



2.32488 
0.497 15 
2.82203 
663.79. 



log 49 = 1.69020 
log ft = 1.16244 

log F = 2.85264 
.-. F = 712.27. 



Hence s' - F = 105.46, and F 



48.48. 



13. Each side of a regular polygon of n sides is c ; show that the radius 
of the circumscribed circle is equal to (c/2) esc (180°/^)? and the radius 
of the inscribed circle is equal to (c/2) cot (180°/w). 

The radius of circumscribed circle = r. 
The radius of inscribed circle = ft. 

C/2 = 180°/w. 
.-. r = (c/2) esc (180°/n). 
ft = (c/2) cot (180°/n). 

14. The radius of a circle is k; show that each side of a regular 
inscribed polygon of n sides is 2 k sin (180°/ w), and that each side of a 
regular circumscribed polygon is 2 k tan (lS0°/n). 

For the inscribed polygon r = k. 
For the circumscribed polygon ft = k. 

C/2 = 180°/^- 
c = 2 r sin (C/2) = 2 k sin (180°/?i) for inscribed polygon, 
c = 2 ft tan (C/2) = 2 k tan (180°/») for circumscribed polygon. 



EXERCISE XVIII 69 

15. The area of a regular polygon of sixteen sides inscribed in a circle 
is 100 sq. in. ; find the area of a regular polygon of fifteen sides inscribed 

in the same circle. 

F = ph/2 = 16 hc/2 = 100. (1) 

C/2 = 180°/ n = 11° 15'. 
c/2 = r sin (C/2). (2) 

h = r cos (C/2). (3) 

From (1), (2), (3), 





4r = 


: Vl00/[sin 


(C/2) cos (C/2)]. 




log 100 = 


: 22.00000 - 


20 


log 


sin ( C/2) = 


: 9.29024 - 


-10 


log 


cos (C/2) = 
.-. log4r = 


: 9.99157- 


-10 




: 2.71819/2 


1 = 1.35909 




.-. r — 


: 22.8605/4 : 


= 5.7151. 



For the polygon of fifteen sides 

C/2 = 180°/n = 12°. 
F = ph/2 = 15 hc/2 =15 r 2 sin (C/2) cos (C/2). 

log 15 = 1.17609 

log(r 2 ) = 2 log r = 1.51405 

log sin (C/2) = 9.31788 - 10 
log cos ( C/2) = 9.99040 - 10 
.-. log F = 1.99842 + 
.-. F= 99.64. 

16. The radius of a circle is 10 ; find the area between the perimeters 
of two regular polygons of thirty-six sides each, one circumscribing the 
circle and the other inscribed in it. 

C/2 = 180°/)i = 5°. 
Let F — area of inscribed polygon, 

and F' = area of circumscribed polygon. 

Then F' — F — area between perimeters of these 

two polygons. 
For inscribed polygon r = 10. 

For circumscribed polygon h = 10. 

F = ph/2 = 36 hc/2 = 36 r 2 sin (C/2) cos (C/2). 
F' = ph/2 = 36 hc/2 = 36 W tan (C/2). 
log 36 = 1.55630 log 36 = 1.55630 

log (h?) = 2. 00000 log r 2 = 2 . 00000 

log tan (C/2) = 8.94195 - 10 log sin (C/2) = 8.94030 - 10 

.-, log F' = 2.49825 log cos (C/2) = 9.99834 - 10 

.-. F'= 314.957. •'• log F = 2.49494 

.-. F'-F= 2.393. .-. F= 312.564. 



70 



PLANE TRIGONOMETRY 



EXERCISE XIX 

Solve each of the following triangles : 
1. Given B = 60° 15', C = 54° 30', a = 100. 
A = 180° - (B + C) = 65° 15'. 
b = a sin B/sin A. 



loga= 2.00000 
log sin B = 9.93862 - 10 
log product = 11.93862 - 10 
log sin A = 9.95815 -10 
.-. log 6= 1.98047 

.-. b = 95.6025. 



a sin C/sin A. 



loga= 2.00000 
log sin C = 9.91069 - 10 
log product = 11.91069 - 10 
log sin A = 9.95815 - 10 
.-. logc= 1.95254 

.-. c = 89.648. 



2. Given A = 45° 41', G = 62° 5 r , b = 100. 
B = 180° - (A + C) = 72° 14'. 
a = 6 sin A /sin 5. 



log 6= 2.00000 
log sin ^4 - 9.85460 - 10 
log product = 11.85460 - 10 
log sin B = 9.97878 - 10 
.-. loga= 1.87582 

.-. a = 75.132. 



c = b sin C/sin B. 



log 6 
log sin C 



2.00000 
9.94627-10 



log product = 11.94627 - 10 
log sin B = 9.97878 - 10 
.-. logc= 1.96749 

.-. c = 92.788. 



3. Given B = 70° 30 r , G = 78° 10', a = 102. 
A = 180° - (B + G) = 31° 20'. 
6 = asin-B/sin-"!. 



loga= 2.00860 
log sin .B = 9.97435 - 10 
log product = 11.98295 - 10 
log sin A = 9.71602 - 10 
.-. log 6= 2.26693 

.-. b = 184.896. 



c = a sin C/sin A. 



loga= 2.00860 
log sin C = 9.99067-10 
log product = 11.99927 - 10 
log sin A = 9.71602 -10 
.-. logc= 2.28325 

.-. c = 191.978. 



EXERCISE XIX 



71 



4. Given A = 55°, B = 65°, c = 270. 

C = 180° -(A + B) = 60°. 

a = c sin A/sin C. 

logc= 2.43136 
log sin A = 9.91336 - 10 
. log product = 12.34472 - 10 
log sin C = 9.93753 -10 
.-. loga = 2.40719 

.-. a = 255.38. 



b = c sin B/sin C. 

\ogc= 2.43136 
log sin B = 9.95728 - 10 
log product = 12.38864 - 10 
log sin C = 9.93753 -10 
.-. log 6= 2.45111 

.-. b = 282.56. 



5. Given a = 123, B = 29° 17', C = 135°. 

A = 180° - (B + G) = 15° 43'. 

b = asin.B/sin.4. 

loga= 2.08991 
log sin B = 9.68942 -10 
. log product = 11.77933 - 10 
log sin ^=9^43278-10 



log b= 2.34655 
,-. 6 = 222.1. 



c = asinC/sin A. 

loga= 2.08991 
log sin C = 9.84949-10 
log product = 11.93940 - 10 
log sin A = 9.43279 - 10 
.-. logc= 2.50661 

.-. c = 321.08. 



6. Given b = 1006.62, A = 44°, C = 70' 
B = 180° - (A + G) = 66°. 
a = b sin A /sin B. 



log 6= 3.00287 
log sin A = 9.84177 -10 
.-. log product = 12.84464 - 10 
log sin B = 9.96073 - 10 
.-. loga= 2.88391 

.-. a = 765.43. 



c = b sin C/sin 5. 

log 6= 3.00287 
log sin C = 9.97299 - 10 
log product = 12.97586 - 10 
log sin B = 9.96073 - 10 
.-. logc= 3.01513 

.-. c = 1035.4. 



7. A ship S can be seen from each of two points A and B on the 
shore. By measurement, AB = 800 ft., Z SAB = 67° 43', and Z SBA 
= 74° 21' 16". Find the distance of the ship from A. 

In the triangle SAB we have given A = 67° 43', B = 74° 21' 16", 
s = 800, to find 6. 

S = 180° - (A + B) = 37° 55' 44 ,/ . 
b = s sin J5/sin S. 



72 PLANE TRIGONOMETRY 

logs= 2.90309 
log sin B = 9.98360 - 10 
.-. log product = 12.88669 - 10 
log sin S = 9.78865-10 
.-. log b= 3.09804 

.-. b = 1253.3. 

8. A flag pole A is observed from two points B and C, 1863 ft. apart. 
Given A BCA = 36° 43' and Z CBA = 57° 21', find the distance of the 
flag pole from the nearer point. 

Here a = 1863, C = 36° 43', B = 57° 21' ; to find c. 

A = 180° - (B + C) = 85° 56'. 



c = asinC/sinJ.. 

loga= 3.27021 
log sin C= 9.77660- 


-10 


.-. log product = 13.04681 - 
log sin A= 9.99891- 


-10 
-10 


.-. logc= 3.04790 
.-. c = 1116.6. 





9. To determine the distance of a hostile fort A from a place J5, a 
line BC and the angles ABC and BCA were measured and found to be 
322.55 yd., 60° 34', and 56° 10' respectively. Find the distance AB. 

Here a = 322.55, B = 60° 34', C = 56° 10'; to find c. 

A = 180° - {B + C) = 63° 16'. 
c = a sin C/sin .A. 

loga= 2.50860 
log sin C = 9.91942 - 10 
.-. log product = 12.42802 - 10 
log sin A = 9.95090 -10 
.-. logc= 2.47712 

.-. c = 300. 

10. A balloon is directly over a straight level road, and between two 
points on the road from which it is observed. The points are 15847 ft. 
apart, and the angles of elevation are found to be 49° 12 x and 53° 29' 
respectively. Find the distance of the balloon from each point of 
observation. 



EXERCISE XIX 73 

Here A = 49° 12', C = 53° 29', b = 15847; to find a and c. 

B = 180° - (J. + C) = 77° 19'. 

a — b sin J./sin B. c = b sin C/sin 5. 

]ogb= 4.19995 log&= 4.19995 

log sin A = 9.87909 - 10 log sin C = 9.90509 -10 

.-. log product = 14.07904 - 10 .-. log product = 14.10504 - 10 

log sin B= 9.98927 -10 log sin B= 9.98927 -10 

.-. loga= 4.08977 .-. log c = 4.11577 

.-. a = 12296. .-. c = 13055. 

11. To find the distance from a point A to a point B on the opposite 
side of a river, a line AC and the angles CAB and ACB were measured 
and found to be 315.32 ft., 58° 43', and 57° 13' respectively. Find the 
distance AB. 

Here A = 58° 43 r , C = 57° 13', b = 315.32 ; to find c. 

B = 180° -(A + C) = 64° 4'. 
c = b sin C/sin 5. 

log&= 2.49875 
log sin C = 9.92465 - 10 
.-. log product = 12.42310 - 10 
log sin B = 9.95391 -10 
.-. logc= 2.46949 

.-. c = 294.77. 

12. From points A and B, at the bow and stern of a ship respectively, 
the foremast C of another ship is observed. The points A and B are 
300 ft. apart; the angles ABC and BAG are found to be 65° 31" and 
110° 46 7 respectively. What is the distance between the points A and C 
of the two ships ? 

Here A = 110° 46 7 , B = 65° 31', c = 300 ; to find b. 
C = 180° - (A + 5) = 3° 43'. 
6 = c sin .B/sin C- 

logc= 2.47712 
log sin 5 = 9.95908-10 
.-. log product = 12.43620 - 10 
log sin C = 8.81173 -10 
.-. log 6= 3.62447 

.-. 6 = 4211.8. 



74 



PLANE TRIGONOMETRY 



EXERCISE XX 

Solve each of the following triangles : 
1. Given a = 145, b = 178, B = 41° 10'. 

Here a < b and B < 90° ; hence A < 90° and there is only one triangle 
having the given parts. 



sin A = u sin B/b. 
loga= 2.16137 
log sin B = 9.81839 - 10 
\ log product = 11.97976 - 10 
log 6= 2.25042 
.-. log sin A= 9.72934-10 
.-. A = 32° 25' 36". 



c = b sin C/sin B. 
log 6= 2.25042 
log sin C = 9.98194 - 10 
log product = 12.23236 - 10 
log sin B = 9.81839-10 
.-. logc= 2.41397 
.-. c = 259.4. 



C = 180° - {A +B) = 106° 24' 24". 
Observe that we use the given side b and its opposite angle B in find- 
ing both A and c. Note the advantage of this in the logarithmic work,' 
especially when there are two solutions. 

2. Given b = 573, c = 394, B = 112° 4'. 

Since B > 90°, A < 90° and C < 90°, and there is only one solution. 



sin C = c sin B/b. 

logc= 2.59550 

log sin B = 9.96696 - 10 

.-. log product = 12.56246 - 10 

log 5= 2.75815 



a = b sin A/sin B. 
log 6= 2.75815 
log sin A = 9.67651 - 10 
.-. log product = 12.43466 - 10 
log sin B = 9.96696-10 
.-. loga= 2.46770 
.-. a = 293.56. 



.-. log sin C= 9.80431-10 

.-. C = 39° 35' 12". 

A = 180° - (B + C) = 28° 20' 48". 

3. Given a = 5.98, b = 3.59, A - 63° 50'. 

Here b < a and A < 90° ; hence B < 90 and there is only one solution. 



sin B = b sin A/a. 


c = a sin C/sin A. 


log 6= 0.55509 
log sin A = 9.95304-10 
log product = 10.50813 - 10 

loga= 0.77670 
.-. log sin B= 9.73143-10 


loga= 0.77670 
log sin C= 9.99726-10 
.-. log product = 10.77396 - 10 
log sin A = 9.95304-10 
.-. logc= 0.82092 


.-. B = 32° 36' 9". 


.-. c = 6.621. 


C = 180° - {A + B) = 


83° 33' 51". 



EXERCISE XX 



75 



4. Given a = 140.5, b = 170.6, A = 40°. 

sin B = b sin A/a. 

log 6= 2.23198 
log sin A = 9.80807 - 10 
.-. log product = 12.04005 - 10 
log a = 2.14768 
.-. log sin B = 9.89237 -10 

.-. B x = 51° 18' 22", B 2 = 128° 41' 38". 
Both values of B satisfy I and II in § 53 ; hence there are two solutions. 



d = 180° - (A + B{) = 88° 41' 38' 

Ci = a sin Ci/sin A. 

\oga= 2.14768 
log sin Ci = 9.99989 - 10 



.-. log product = 12.14757 
log sin A= 9.80807 



10 
10 





.-. lo 


gCi 


= 2.33950 








Cl 


= 218.53. 




5. 


Given 


b = 


74.1, c= 64.2, C = 

sinB = 

' log 6 = 
log sin C = 


27° 18'. 

b sin C'/c. 

1.86982 
9.66148 



C 2 = 180° - (A + B 2 ) = 11° 18' 22' 

c 2 = a sin C2/sin .4. 

loga= 2.14768 
log sin C 2 = 9.29237 - 10 
.-. log product = 11.44005 - 10 
log sin A = 9.80807 -10 
logc 2 = 1.63198 

.-. c 2 = 42.853. 



10 



log product = 11.53130 
los:c= 1.80754 



10 



.-. log sin B= 9.72376 -10 

.-. By. = 31° 6T 46", B 2 = 148° 2 r 14". 

Both values of B satisfy I and II in § 53 ; hence there are two solutions, 

A x = 180° - (C+Bi) = 120° 44' 14". 

a\ = c sin Ai/sm C. 

\ogc= 1.80754 
log sin A x = 9.93425 - 10 
.-. log product = 11.74179 - 10 
log sin C = 9.66148 - 10 
.-. logai = 2.08031 

.-. ai = 120.31. 



= 180° - (C + B 2 ) = 4° 


39' 46 


a 2 = 


csinJ. 2 /sinC. 


logc = 


1.80754 




log sin A 2 = 


8.91004 


-10 


log product = 


10.71758 


-10 


log sin C — 


9.66148 


-10 


.-. loga 2 = 


1.05610 




.-. a 2 = 


11.379. 





76 PLANE TRIGONOMETRY 

6. Given a = 27.89, b = 22.71, B = 65° 38'. 

sin A = a sin B/b. 

log a = 1.44545 
log sin B = 9.95948 - 10 
.-. log product = 11.40493 - 10 
log 5= 1.35622 
.-. log smA = 10.04871 -10 

Here log sin A is +, or sin A > 1, which is impossible; hence the tri- 
angle is impossible. 

7. Given b = 45.21, c = 50.3, B = 40° 32' 7". 

sin C = c sin B/b. 

logc= 1.70157 
log sin b = 9.81286 -10 
.: log product = 11.51443 - 10 

log 6 = 1.65523 
.-. log sin C = 9.85920 - 10 

.-. Ci = 46° 18' 40", C 2 = 133° 41' 20". 

Both values of C satisfy I and II in § 53; hence there are two solutions. 

A x = 180° - (B + Ci) = 93° 9' 13". A 2 = 180° - (B + C 2 ) = 5° 46' 33". 

ai = b sin Ai/sin B. a 2 = 6 sin A 2 /sin B. 

log 6= 1.65523 log 6= 1.65523 

log sin A x = 9.99934-10 log sin A 2 = 9.00276 -10 

.-. log product - 11.65457 - 10 .-. log product = 10.65799 - 10 

log sin B = 9.81286 - 10 log sin B = 9.81286 - 10 

.-. logai= 1.84171 .-. loga 2 = 0.84513 

.-. a; = 69.457. .-. a 2 = 7.0005. 

8. Given a = 34, b = 22, B = 30° 20'. 

sin A = a sin B/b. 

\oga= 1.53148 
log sin B = 9.70332 -10 
.-. log product = 11.23480 - 10 
log b = 1.34242 

.-. log sin A = 9.89238 - 10 

.-. Ax = 51° 18' 27", A 2 = 128° 41' 33". 



EXERCISE XX 



77 



Both values of A satisfy I and II in § 53 
d = 180° - (B + A 2 ) = 98° 21' 33". C 2 
Ci = 6 sin Ci/sin B. 
log 6= 1.34242 
log sin d = 9.99536 - 10 
.-. log product = 11.33778 - 10 
log sin B = 9.70332-10 
.•'. logCi= 1.63446 
.-. ci = 43.098. 



hence there are two solutions. 
= 180° - (B + vi 2 ) = 20° 58' 27 // . 
c 2 = b sin C 2 /sin jB. 
log&= 1.34242 
log sin C 2 = 9.55382 - 10 
log product = 10.89624 - 10 
log sin B = 9.70332 - 10 
.-. logc 2 = 1.19292 
.-. c 2 = 15.593. 



9. Given a = 55.55, b = 66.66, B = 77° 44' 40". 
Here a < 6, and 5 < 90°; hence -4 < 90°, and there is only one solution. 



sin A 



a sin B/b. 
loga= 1.74468 
log sin B = 9.98999-10 
.-. log product = 11.73467 - 10 
log b= 1.82387 
.-. log sin A = 9.91080 -10 
.-. A = 54° 31' 13". 
C = 180° - (A + B) = 47° 44' 7". 

309, b = 360, A = 21° 14' 25' 
sin B = b sin A /a. 
log 6= 2.55630 
log sin ^ = 9.55904 - 
.-. log product 



c = b sin C/sin B. 

log 6= 1.82387 

log sin C = 9.86925 -10 

log product = 11.69312 - 10 

log sin B= 9.98999-10 



logc= 1.70313 
.-. c = 50.481. 



10. Given a 



10 



12.11534 -10 
log a = 2.48996 
.-. log sin B = 9.62538 - 10 
.-. B x = 24° 57' 54", B 2 



155° 2' 6' 



Both values of B satisfy I and II in § 53; hence there are two solutions. 



d = 180° -{A+ B x ) = 133° 47' 4V 

Ci = a sin Ci/sinA. 



\oga= 2.48996 

log sin d = 9.85843 

log product = 12.34839 

log sin A = 9.55904 

.-. logCi= 2.78935 

.-. ci = 615.67. 



= 180° - (A + B 2 ) = 3° 43' 29". 
c 2 = a sin C 2 /sin A. 
log a = 2.48996 
log sin C 2 = 8.81267 - 10 
log product = 11.30263 - 10 
log sin A = 9.55904-10 
.-. logc 2 = 1.74359 
.-. c 2 = 55.41. 



78 



PLANE TRIGONOMETRY 



11. 



Given b = 19, c = 18, C = 15° 49'. 

sinB = b sin C/c. 



log& 
log sin C 



1.27875 
9.43546 



10 



log product = 10.71421 

log c = 1.25527 

.-. log sin B 



10 



.45894 - 10 
.-. B x = 16° 43' 13", B 2 = 163° 16' 47". 
Both values of jB satisfy I and II in § 53 ; hence there are two solutions. 

A 2 = 180° - (C + B 2 ) = 0°54 / 13". 
a 2 = csinJ. 2 /sinC. 



A 1 =1S0° - (C+B{) =147°27 / 47 
«i = c sin ^i/sin C. 



logc= 1.25527 
logsin^4i= 9.73065- 


-10 


>g product = 10.98592 - 
log sin C = 9.43546 - 


- 10 
-10 


.-. log«i= 1.55046 
.-. ai= 35.519. 





logc= 1.25527 
log sin A 2 = 8.19784-10 
log product = 9.45311 - 10 
log sin C = 9.43546 - 10 
.-. log a 2 = 0.01765 

.-. a 2 = 1.0415. 



EXERCISE XXI 

Solve each of the following triangles : 
1. Given a = 266, b = 352, C = 73°. 



B-A b-a C 

tan — — = cot — . 

2 b + a 2 


c = asin C/sin A. 


Here b + a = 618, b - c = 86, C/2 = 


-. 36° 30'. 


log (6- a) = 1.93450 


loga= 2.42488 


log cot (C/2) = 0.13079 


log sin C= 9.98060-10 


.-. log product = 12.06529-10 


.-. log product = 12.40548-10 


log (6 + a) = 2.79099 


log sin A= 9.83255-10 


.-. log tan l(B-A)= 9.27430 - 10 


.-. logc = 2.57293 


.-. (B -A)/2 =10°39 / 3". 


.-. c= 374.05. 



(B + A)/2 = 90° - C/2 = 53° 30' 

.-. A = 42° 50' 57". 
B = 64° 9' 3". 



EXERCISE XXI 



79 



2. Given b 



tan 



91.7, c 
B- C 



31.2, A 
b-c 



33° 7' 9' 



A 

— cot — 

b + c 2 



a = 6 sin .A/sin 5. 



Here 



6 + c = 122 

log (b - c) 
log cot (A/2) 



9, 6 



60.5, .4/2 = 16° 33' 35' 



1.78176 
0.52674 



.-. log product =2.30850 
log (b + c) = 2.08955 
log tan $(B-C)= 0.21895 
.-. (B - C)/2 = 58° 52' 2". 
(B + C)/2 = 90° - A/2 
.-. 5 = 132° 18' 27' 
C = 14° 34' 23". 



log 6 

log sin A 

.-. log product 

log sin B 

.-. log a 

.-. a 

73° 26' 25". 



1.96237 

9.73750-10 
11.69987-10 

9.86897-10 

1.83091 
67.75. 



3. Given a = 960, 

A 

tan 



5 = 720, C 
- B a-b 



cot 

a + b 2 



25° 40'. 
C 



2 

Here a + 6 = 1680, a - b = 240 
log (a -b) = 2.38021 
logcot(C/2) = 0.64243 



c = a sin O/sin A. 



.-. log product = 13.02264-10 
log (a + b) = 3.22531 
log tan i ( J. - B) = 9. 79733 - 10 
.-. (A - B)/2 = 32° 5' 30". 
[A + J5)/2 = 90° - C/2 = 
.-. ^4 = 109° 15' 30". 
B = 45° 4' 30". 



C/2 

log a 

log sin C 

log product 

log sin A 

.-. logc 

.-. c 



12° 50'. 
2.98227 
9.63662-10 



12.61889- 
9.97499- 



2.64390 
440.45. 



7° 10'. 



4. Given a 



886, b 

; J.-J5 

tan 

2 

Here a + b = 1633, a 

log (a-b) = 

log cot (C/2) = 

.-. log product = 

log (a + b) = 

.-. log tan | (J. -5) = 

.-. (A - B)/2 = 



747, C 

a-6 C 

cot - ■ 

a + b 2 



1°54' 



- b = 139, 

2.14301 

0.13954 



12.28255- 
3.21299 



10 



C/2 = 35° 57'. 

log a 
log sin C 
.-. log product 
log sin A 



9.06956-10 
41' 39". 
(A + B)/2 = 90° - C/2 = 54 c 
.-. A = 60° 44' 39". 
B = 47° 21 21". 



logc 
.-. c 



a sin C/sin A. 



2.94743 

9.97796-10 
12.92539-10 

9.94074-10 

2.98465 
965.28. 



80 



PLANE TRIGONOMETKY 



Given b = 41.02, c = 45 
C - B c- 



),A = 62° 9' 38* 



tan 



c-6 ^ 
cot — • 

c + b 2 



a = csin^l/sinC. 



Here c + b = 86.51, c - 6 = 4.47, .4/2 = 31° 4' 49' 



log (c - b) = 0.65031 

log cot (^4/2) = 0.21985 

.-.log product = 10.87016- 

log (c + 6) = 1.93707 

.-. logtani(C- J B)= 8.93309- 

.-. (C - B)/2 = 4° 53' 58" 



10 



10 



logc= 1.65792 

log sin A = 9.94658-10 

log product = 11.60450-10 

log sin C = 9.95299-10 

.-. loga = 1.65151 

.-. a = 44.824. 



(C + B)/2 = 90° - A/2 = 58° 55' ll". 
.-. B = 54° 1' 13". 
C = 63° 49' 9". 

6. Two trees A and B are on opposite sides of the pond. The dis- 
tance of A from a point C is 297.6 ft., the distance of B from C is 
864.4 ft., the angle ACB is 87° 43' 12". Find the distance AB. 

Here b = 297.6, a = 864.4, C = 87° 43' 12" ; to find c. 



A-B a-b C 
tan = cot — • 

2 a + 6 2 


c = 6 sin C/sin #. 


a + 6 = 1162, a - 6 = 


= 566.8, C/2 = 43° 51' 36". 


log (a -6) = 2.75343 


log 6= 2.47363 


log cot (C/2) = 0.01729 


log sin C = 9.99966-10 


.-. log product = 12.77072-10 


.-. log product = 12.47329-10 


log (a + 6) = 3.06521 


log sin B= 9.51763-10 


.-. log tan i (^4-5)= 9.70551-10 


.-. logc= 2.95566 


.-. (A - B)/2 = 26° 54' 43". 


.-. c = 902.94. 


(^4 + B)/2 = 90° - C/2 = 46° 


8' 24". 


.-. A = 73° 3' 7". 


- 


5 = 19° 13' 41". 





7. Two mountains A and B are respectively 9 and 13 mi. from a 
town C, and the angle ACB is 71° 36' 37". Find the distance between 
the mountains. 

Here 6=9, a = 13 ; C = 71° 36' 37" ; to find c. 



tan- 



A-B 



a-b a 

■ cot - • 

a + b 2 



a + 6 = 22, a- 6 = 4, C/2 



c = 6 sin C/sin B. 
35° 48' 19". 



EXERCISE XXI 



81 



log (a - b) 
•cot(C/2) 



0.60206 
0.14184 



.-. log product = 10.74390-10 .-. 1 

log (a + b) = 1.34242 
logtani(A-B)= 9.40148-10 

.-. (A-B)/2 =14°8 / 48 // . 
(A + B)/2 = 90° - C/2 = 54° 11' 41' 

.-. A = 68° 20' 29 // . " 
E = 40° 2' 53". 



logb = 0.95424 

log sin C = 9.97723-10 

^ product = 10.93147-10 

log sin B = 9.80850-10 

.-. logc = 1.12297 

.-. c = 13.273. 



8. Two points A and B are visible from a third point C, but not from 
each other. The distances AC, BC, and the angle ACB were measured 
and found to be 1321 ft., 1287 ft., and 61° 22' respectively. Find the 
distance AB. 



Here b = 1321, a = 1287, C = 61° 22'; to find c. 



tan 



B 



a C 
— cot — ' 
a 2 



b sin C/sin B. 



b + a = 2608, b - a = 34, C/2 = 30° 41'. 



log (6- a) = 1.53148 
log cot ( C/2) = 0.22668 



3.12090 
9.94335-10 



.-. log product = 11.75816-10 
log (b + a) = 3.41631 
log tan* (.8-^)= 8.34185-10 

(B -A)/2 = 1°15 / 31 ,/ . 

(B + A)/2 = 90° - C/2 = 59° 19'. 

.-. A = 57° 3' 29". 
B = 60° 34' 31". 



log b 

log sin C 

log product = 13.06425- 

log sin B = 9.94002- 

.-. logc= 3.12423 

.-. c = 1331.2. 



9. From a point 3 mi. from one end of an island and 7 mi. from the 
other end, the island subtends an angle of 33° 55 r 15". Find the length 
of the island. 

Let A and B be the ends of the island, and C the point of observation ; 
then we have a = 7, b = 3, C = 33° 55' 15"; to find c. 



tan 



A-B a-b C 
= cot — • 

2 a + b 2 




a + b = 10, a - b = 4, C/2 = 


= 16° 57' 



c = b sin C/sin B. 



82 PLANE TRIGONOMETRY 

log (a - 6) = 0.60206 log b = 0.47712 

log cot (0/2) = 0.51573 log sin C = 9.74667-10 

.-. log product = 1.11779 .-. log product = 10.22379-10 

log (a + b) = 1.00000 logsin£ = 9.54154-10 

.-. log tan i (A- B) = 0.11779 .-. log c = 0.68225 

.-. {A - B)/2 = 52° 40' 35". .-. c = 4.8111. 

(A + B)/2 = 90° - 0/2 = 73° 2' 22 // . 

.-. A = 125° 42' 57 ,/ . 
B = 20° 21' 47". 

10. Two stations J. and B on opposite sides of a mountain are both 
visible from a third station O. The distances AC and BC and the angle 
^4Cj5 were measured and found to be 11.5 mi., 9.4 mi., and 59° 31' 
respectively. Find the distance from A to B. 

Here 6 = 11.5, a = 9.4, C = 59° 31'; to find c. 

B-A b-a C _ . n . * 

tan = cot - • c = o sin (7/sin B. 

2 6 + a 2 

6 + a = 20.9, & - a := 2.1, 0/2 = 29° 45' 30". 

log (6- a) = 0.32222 log 6= 1.06070 

log cot (0/2) = 0.24280 log sin O = 9.93539-10 

.-. log product = 10.56502-10 .-. log product = 10.99609-10 

log (b + a) = 1.32015 log sin B = 9.97356-10 

.-. log tan i(B-A)= 9.24487-10 .-. logc= 1.02253 

.-. (B - A)/2 = 9° 68' 2". .-. c = 10.532. 

(B + A)/ 2 = 90° - 0/2 = 60° 14' 30". 

.-. A = 50° 16' 28". 
B = 70° 12' 32". 

11. Two trains leave the same station at the same time on straight 
tracks that form an angle of 21° 12'. Their average speeds are 40 mi. 
and 50 mi. an hour respectively. How far apart will they be at the end 
of the first thirty minutes ? 

Let O be the station and A and B the trains after thirty minutes ; then 

we have b = 20, a = 25, O = 21° 12 7 ; to find c. 

b sin O/sin B. 



tan 


A-B _ 

2 


a-b O 

= cot - . 

a + b 2 








c = 




a + b -- 


= 45, a — b = 


5, 


0/2 = 


= 10° 


36 7 . 



EXERCISE XXII 83 



log (a -6)= 0.69897 


log 6= 1.30103 




log cot (C/2) = 0.72782 


log sin C = 9.55826- 


-10 


.-. log product = 11.42679-10 


.-. log product = 10.85929- 


-10 


log(a + &) = 1.65321 


logsinB= 9.87580- 


-10 


log tan i (A- B) = 9.77358-10 


.-. logc = 0.98349 




.-. (A - B)/2 = 30° 41' 54". 


/. c = 9.627. 




(A + B)/2 = 90° - C/2 = 


19° 24'. 




.-. A = 110° 5' 54". 






5 = 48° 42' 6". 







EXERCISE XXII 

Solve each of the following triangles : 

1. Given a = 56, b = 43, c = 49 ; to find A, B, C. 



Formulas 



r = V(s — a) (s — b) (s — c)/s, 
tan (A/2) — r/(s — a), 
I tan (B/2) = r/(s - 6), 
[tan (C/2) = r/(s- c). 

Here s = 74, s - a = 18, s - 6 = 31, s - c = 25. 

log (s - a) = 1.25527 .-. log tan (4/2) = 9.88240 - 10. 

log (s - b) = 1.49136 log tan (2?/2) = 9.64631 - 10. 

log(s - c) = 1.39794 log tan (C/2) = 9.73973 - 10. 

.-. log product = 4.14457 .-. A/2 = 37° 20' 9". 

log s = 1.86923 B/2 = 23° 53' 19" 

.-. logr = 2.27534/2 C/2 = 28° 46' 32". 

= 11. 13767 - 10. .-. A = 74° 40' 18". 

Check. A + B + C = 180°. £ = 47° 46' 38". 

C = 57° 33' 4". 
2. Given a = 8.5, b = 9.2, c = 7.8. 

Here s = 12.75, s - a = 4.25, s - b = 3.55, s - c = 4.95. 

log(s- a) =0.62839 .-. log tan (A/2) = 9.75547 - 10. 

log(s - 6) = 0.55023 log tan B/2) = 9.83363 - 10. 

log (s-c)= 0.69461 log tar ^/2) = 9.68925 - 10. 

.-. log product = 1.87323 A/2 = 29° 39' 37". 

log s = 1.10551 B/2 = 34° 1 V 4". 

.-. log r = 0. 76772/2 C/2 = 26° 3' 20.6". 

= 10.38386-10. A = 59° 19' 14". 

C^ecA:. 4 + 5 + = 180°. 5 = 68° 34' 8". 

C = 52° 6' 41". 



84 



PLANE TRIGONOMETRY 



3. Given a = 61.3, b = 84.7, c 


= 47.6. 




Here s = 96.8, s — a = 35.5, s - 


-6 = 12.1, s- c = 49.2. 




log(s- a) = 1.55023 


.-. logtan(J./2) = 9.619325 - 


10. 


log(s-&) = 1.08279 


log tan (B/2) = 0.086765. 




log(s - c) = 1.69197 


logtan(C/2) = 9.477585 - 


10. 


. log product =4.32499 


.-. A/2 = 22° 35' 54" 




logs = 1.98588 


B/2 = 50° 41' 8". 




.-. logr = 2.33911/2 


(7/2 = 16° 42' 58" 




= 11.169555-10. 


A = 45° 11' 48" 




Check. A + B + C = 180°. 


B = 101° 22' 16' 
C = 33° 25' 56" 




4. Given a = 705, 6 = 562, c = 


639. 




Here s = 953, s- a = 248, s - 


b = 391, s-c = 314. 




log(s- a) = 2.39445 


.-. log tan {A/2) = 9.857785 - 


10. 


log(s - b) =2.59218 


log tan (B/2) = 9.660055 - 


10. 


log(s- c)= 2.49693 


log tan ((7/2) = 9.755305 - 


10. 


. log product = 7.48356 


.-. A/2 = 35° 46' 56.5". 


logs = 2.97909 


B/2 = 24° 34' 2.6' 




.-. log r = 4.50447/2 


(7/2 = 29° 39' 3". 




= 12.252235 - 10. 


.-. A = 71° 33' 53" 
5 = 49° 8' 5". 

C = 59° 18' 6". 




5. Given a = .0291, b = .0184, 


c = .0358. 




Here s = .04165, s - a = .01255, s - b = .02325, s - c = .00585. 




log(s-a)= 8.09864-10 


.-. log tan (A/2)= 9.70766- 


10. 


log(s-6)= 8.36642-10 


logtan(5/2)= 9.43988- 


10. 


log(s-c)= 7.76716-10 


logtan(0/2) = 10.03914- 


10. 


. log product = 24.23222 -30 


.-. A/2 = 27° 1' 35". 




logs= 8.61962-10 


B/2 = 15° 23' 41" 
12 (7/2 = 47° 34' 42" 




.-. logr =(15.61260 -20), 




= 17.80630 - 20. 


.-. A = 54° 3' 10". 
B = 30° 47' 22" 
C = 95° 9' 24". 




6. Given a = 85, b = 127, A = 


26° 26'. 




log b = 


2.10380 




log sin A = 


9.64851 - 10 




.-. log product = 


11.75231 -10 




loga = 


1.92942 




.-. log sin B = 


9.82289 - 10 




.-. B x = 


41° 41' 26", B 2 = 138° 18' 34". 





EXERCISE XXII 



85 



Both values of B satisfy I and II in § 53 ; hence there are two solutions. 



d = 180° -(A + Bi) = 


lll°52 / 34 // . 






C 2 = 180° - (A + B 2 ) = 


15° 15' 26". 






Ci = asm Ci/sin A. 


c 2 = 


a sin C 2 /sin J. 


loga= 1.92942 


loga = 


1.92942 




log sin d = 9.96754-10 


log sin C2 = 


9.42021 - 


-10 


log product = 11.89696 - 10 


log product = 


11.34963 - 


-10 


log sin A= 9.64851 -10 


log sin A = 


9.64851 - 


-10 


.-. logCi= 2.24845 


.-. log c 2 = 


1.70112 




.-. a = 177.2. 


.-. c 2 = 


50.248. 




7. Given a = 5.953, b = 9.639, C = 134 









B -A b-a C 
tan = cot — • 

2 b + a 2 


c = 


asinC/sin A. 


6 + a = 15.592, b - a = 


= 3.686, C/2 = 


67°. 




log (6- a) = 0.56656 


loga = 


0.77474 




logcot(C/2) = 9.62785 - 10 


log sin C = 


9.85693 - 


-10 


.-. log product = 10. 19441 - 10 


log product = 


10.63167 - 


-10 


log(& + a) = 1.19290 


log sin A = 


9.47257 - 


-10 


Log tan \ (B- A) = 9.00151-10 


.-. log c = 


1.15910 




.-. {B - A)/2 = 5° 43 7 49 // . 


.-. c = 


14.424. 




(B + A)/2 = 90° - C/2 = 23°. 








.-. A = 17° 16' 11". 








B = 28° 43 y 49 ,/ . 









8. Given a = 3019, b = 6731, c = 4228. 



Here s = 6989, 



a = 3970, 



log(s- a) = 3.59879 

log(s- 6) = 2.41162 

log {s-c) = 3.44107 

. log product = 9.45148 

log s = 3.84442 

.-. logr= 5.60706/2 

= 12.80353-10. 
Check. A + B + C = 180°. 



258, s-c = 2761. 

log tan (A/2) = 9.20474 - 10. 
log tan (B/2) = 0.39191 - 10. 
log tan (C/2) = 9.36246 - 10. 
.-. A/2 = 9° 6' 11". 
B/2 = 67° 55 7 23". 
C/2 = 12° 58 / 26". 
.-. A = 18° 12' 22". 
B = 135° 50' 46". 
C = 25° 56' 52". 



86 



PLANE TRIGONOMETRY 



9. Given a = 60.935, c = 76.097, A = 133° 41', compute C, B, and b. 
Since c> a, C > A > 90°, which is impossible, as a triangle cannot 
have two obtuse angles; hence the triangle is impossible. 



10. Given b = 74. 



738, C = 81° 47'. 



sin B = b sin C/c. 


a = 


c sin ^1/sin 


C. 


log b= 1.87394 


logc = 


1.99448 




log sin C = 9.99552-10 


log sin A = 


9.88194 - 


10 


.-. log product = 11.86946 - 10 


.-. log product = 


11.87642 - 


10 


logc = 1.99448 


log sin C = 


9.99552 - 


10 


.-. log sin B= 9.87498-10 


.-. log a = 


1.88090 




.-. B = 48° 34' 44". 


.-. a = 


76.015. 




A = 180° - (B + C) = 49 


° 38' 16". 






11. Given b = 129.21, c = 28.63, A = 


: 27° IS'. 






B-C b^c A 
tan = cot — 

2 b + c 2 


a = 


c sin A/sin 


0. 


6 + c = 157.84, 6- c 


= 100.58, ^4/2 = 


13° 36' 30" 




log (b- c) = 2.00251 


logc = 


1.45682 




log cot (.4/2)= 0.61604 


log sin A = 


9.66025 - 


10 


.-. log product = 12.61855 - 10 


.-. log product = 


11.11707 - 


10 


log (6 + c)= 2.19822 


log sin C = 


9.09767 - 


10 


log tan |(J5 - C) = 10.42033 - 10 


.: \oga = 


2.01940 




.-. (B - C)/2 = 69° 11' 54". 


.-. a = 


104.57. 




(B + C)/2 = 90° - J./2 = 76° 


23' 30". 






.-. B = 145° 35' 24". 








C = 7° ir 36". 








12. Given a = 2.51, b = 2.79, c = 2.3 


3. 






Here s = 3.815, s - a = 1.305, s - 6 = 


= 1.025, b-c = 


1.485. 




log(s-a)= 0.11561 


log tan (A/2) = 


9.742675 - 


10. 


log(s-6)= 0.01072 


log tan (B/2) = 


9.847565 - 


10. 


log(s-c)= 0.17173 


log tan (C/2) = 


9.686555 - 


10. 


.-. log product = 10.29806 - 10 


.-. A/2 = 


28° 56' 23" 




logs= 0.58149 


B/2 = 


35° 8' 42.7 




.-. logr = (19.71657 - 20)/2 


(7/2 = 


25° 54' 55" 




= 9.858285 -10 


.-. A = 


: 57° 52' 46" 




heck. A + B+C =180°. 


B = 

C = 


70° 17' 25" 
■ 51° 49' 50" 





EXERCISE XXII 



87 



13. Given a = 32.163, c = 27.083, C = 52° 24' 16". 
sin A = a sin C/c. 
loga = 1.50736 
log sin C = 9.89891 - 10 
.-. log product = 11.40627 - 10 
log c = 1.43270 
.-. log sin A = 9.97357 -10 

.-. A x = 70° 12' 48", A 2 = 109° 47' 12". 
Both values of A satisfy I and II in § 53 ; hence there are two solutions. 

= 180° -(C + A 2 ) = 17° 48' 32". 
= c sin J5 2 /sin C. 

logc= 1.43270 
log sin B 2 = 9.48550 - 10 
log product = 10.91820 - 10 
log sin C = 9.89891 -10 
.-. log& 2 = 1.01929 
.-. b 2 = 10.454. 



5i = 180°-(C + ^ 1 ) = 57° 


22' 56". 


B 2 


&i = c sin jBi/sin C. 




h 


logc= 1.43270 






logsin.Bi = 9.92545 


-10 




.-. log product = 11.35815 


-10 




log sin C= 9.89891 


-10 





14. 



.-. log6i= 1.45924 
.-. & 1 = 28.79. 

Given a = 74.8, c = 124.09, B = 83° 26' 52 
C-A c 



tan 



a B 
cot — 

c + a 2 



asinB/sinA. 



c + a- 198.89, c - 
logic -a)= 1.69276 
log cot (B/2) = 0.04977 
.-. log product = 11.74253-10 
log (c + a) = 2.29861 
\ logtani(C - -4) = 9.44392-10 
.-. (C-J.)/2 = 15°31 / 54". 
(C + ^)/2 = 90°-B/2 = 4 
.-. J. = 32° 44' 40". 
C = 63° 48' 28". 

15. Given a = 86.062, c = 63.576, 
sin C = c sin A/a. 
\ogc= 1.80329 
log sin A = 9.51728-10 
.-. log product = 11.32057-10 
log a = 1.93481 
.-. log sin C 



a = 49.29, B/2 = 41° 43' 26". 
loga= 1.87390 
log sin B= 9.99716-10 



log product = 11.87106 - 10 
log sin A= 9.73311 -10 



log&= 2.13795 
.-. 6 = 137.39. 



1° W 54' 



10 



9.38576 
C = 14°4 / 7". 
B = 90° - (A + C) = 146° 43' 10 



4 = 19° 12' 43". 

6 = asin.B/sin-4. 
loga= 1.93481 
log sin B = 9.73937 -10 
.-. log product = 11.67418 - 10 
log sin A = 9.51728 -10 
.-. log 5= 2.15690 
.-. 6 = 143.52. 



88 



PLANE TRIGONOMETRY 



16. Given a - 93.272, b = 81.512, C = 58°. 



A-B a-b C 
tan = cot — • 

2 a + b 2 


c — b sin C/sin B. 


a + b = 174.784, a- 


-6 = 11.76, C/2 = 29°. 


\og(a~b)= 1.07041 


log& = 1.91122 


log cot ( C/2) = 0.25625 


log sin C = 9.92842 - 10 


.-. log product = 11.32666-10 


.-. log product = 11.83964 - 10 


log(a + 6)= 2.24250 


log sin B= 9.90840-10 


.-. log tan i (J. -B) = 9.08416-10 


.-. logc= 1.93125 


.-. (A - B)/2 = 6° 55 / 15". 


.-. c = 85.36. 


(A + B)/2 = 90° - (7/2 = 61°. 




.-. A = 67° 55' 15". 




B = 54° 4' 45". 





17. The sides of a triangular field are 534 ft., 679.47 ft., and 474.5 ft. 
Find the angles. 

Here a = 534, b = 679.47, c = 474.5; to find A, B, C. 

s = 843.99, s-a = 306.99, s - 6 = 164.52, s-c = 369.49. 



log(s- a) = 2.49135 

log(s- b) = 2.21622 

log (s-c) = 2.56760 

log product = 7.27517 

log s = 2.92634 

logr 



.-. log tan (^1/2) = 9.68307 - 10. 
log tan (B/2) = 9.95820 - 10. 
log tan (C/2) = 9.60682 - 10. 
.-. A/2 = 25° 44 y V. 
B/2 = 42° 14' 48". 
C/2 = 22° V 8". 
.-. A = 51° 28 r 14". 
B = 84° 29' 36". 
C = 44° 2 > 16". 

18. A pole 13 ft. long is placed 6 ft. from the base of an embankment, 
and reaches 8 ft. up its face. Find the slope of the embankment. 
Here a = 13, b = 6, c = 8 ; to find 180° - A. 
A l(s -b)(s- c) 



= 4.34883/2 

= 12.17442-10. 



tan 



V* 



s(s — a) 

s = 13.5, s — a — 0.5, s - b = 7.5, s - c = 5.5. 
log(s-6)= 0.87506 logs= 1.13033 

log(3-c)= 0.74036 log (s- a) = 9.69897-10 

.-. log numerator = 11.61542-10 .-.log denominator = 10.82930 - 10 

.39306. 



log tan- =0.78612/2 

.-. A 

180° - A 



(67° 58' 32") 
44° 2' 56". 



135° 57' 4' 



EXERCISE XXII 89 

19. A point P is 13581 in. from one end of a wall 12342 in. long, and 
10025 in. from the other end. What angle does the wall subtend at the 
point P ? 

Here a = 10025, b = 13581, p = 12342 ; to find Z P in A PAB. 
. P 1(8- a) (s - b) 











2 






ab 










s = 


17974, s - 


a 


= ' 


F949, 


s-b = 


-- 4393. 






log(s 


-a) 


= 3.90031 










loga = 


= 4.00108 




log(s 


-&) 


= 3.64276 










log& = 


= 4.13293 


.-. log 


numerator 


= 17.54307 


— 


10 




. log denominator = 


= 8.13401 



.-. logsin(P/2) = (19.40906 - 20)/2 = 9.70453 - 10. 
.-. P = (30° 25' 38") • 2 = 60° 61' W. 

20. The distances between three cities A, B, and C are as follows 
AB — 165 mj., AC = 72 mi., and BC = 185 mi. B is due east from A. 
In what direction is C from A ? 

Here a = 185, b = 72, c = 165; to find ^1. 
^4 _ /s (s — a) 



cos 



J* 



2 X be 

s = 211 s — a = 26. 
logs= 2.32428 ' log& = 1.85733 

log(s-a)= 1.41497 logc = 2.21748 

.-. log numerator = 13.73925 — 10 .-. log denominator = 4.07481 
.-. logcos(^l/2)= (19.66444 - 20)/2 = 9.83222 - 10. 
.-. A = (47° 11' 30 ,/ ) • 2 = 94° 23 / . 
Hence C is N. 4° 23' W., or S. 4° 23' W., from A. 

21. Under what visual angle is an object 7 ft. long seen when the eye 
of the observer is 5 ft. from one end of the object and 8 ft. from the 
other end? 

Here a = 7, 6 = 5, c = 8; to find A. 

A ls(s — a) 



cos 



^ 



2 \ be 

s = 10, s - a = 3. 

logs= 1.00000 log 6 = 0.69897 

log(s-a)= 0.47712 logc = 0.90309 

numerator = 11.47712 — 10 .-. log denominator = 1.60206 
ogcos(^L/2)= (19.87506 - 20)/2 = 9.93753 - 10. 

/. A = 30° x 2 = 60°. 



90 PLANE TRIGONOMETRY 

22. Prove sin A = 2 Vs(s - a)(s - b)(s - c)/(bc). 
sin A = 2 sin {A/2) cos (4/2) 

= 2 J(s-fr)(s-c) J s(s-g) 
\ be \ be 



= 2 Vs (s -a)(s- b) (s - c)/(bc). 

23. Prove cos A = s (s — a)/(6c) — (s — b) (s — c)/(bc). 

cos J. = cos 2 (4/2) - sin 2 (4/2) 

= [s (s - a) - (s - b) (s - c)]/(bc). 

24. If a = 18, 6 = 24, c = 30, show that sin 4 = 3/5. 



By example 22, sin 4 = 2 V s (s — a) (s — 6) (s — c)/(bc). 
Here s = 36, s - a = 18, s - 6 = 12, s - c = 6. 



.-. sin 4 = 2V6-6-6-3-6-2- 6/(6 • 4 • 5 • 6) 
= 2 ■ 6 • 6 • 6/(6 -4-5.6) = 3/5. 

25. TAe sides o/ a triangle can be substituted for the sines of their oppo- 
site angles, and vice versa, when they are involved homogeneously in the 
numerator and denominator of a fraction, or in both members of an equation. 

For this substitution is the multiplication of each term by the equal 
numbers, a/sin 4, 6/sin.B, c/sin C, or their reciprocals. 

E.g., sin 4 = sin (180° - A) = sin (B + C). 

.-. sin 4 = sin B cos C + sin C cos S. by [7] 

Multiplying the first term by a/sin 4, the second by 6/sin J5, and the 
third by c/sin C, we obtain 

a = b cos C -j- c cos B. 1 

Similarly, 6 = acosC + ccos4, > (1) 

and c = acosjB + 6cos4. J 

„ „ 2c 2 + a 2 2sin 2 C + sin 2 4 

26. Prove = ■• (1) 

3a6c 3 b sin 4 sin C 

sin 2 C/c 2 = sin 2 4/a 2 = sin 4 sin C/(ac). 

Multiplying 2 c 2 by sin 2 C/c 2 , a 2 by sin 2 4/a 2 , and 3 abc by sin 4 sin C/(ac), 
we obtain the second member of (1) from its first member. 



EXERCISE XXIII 



91 



27. Multiplying the equations in (1) of example 25 by — a, 6, and c 
respectively, and adding, we obtain the law of cosines, 

b 2 + c 2 - a 2 = 2 be cos A. 

From the relations (1) in example 25 we obtain 

- a 2 = — ab cos C — ac cos B, 

b 2 = ab cos C + be cos ^4, 

c 2 = ac cos U + be cos -4. 

Adding these equations, member to member, we obtain 
b 2 + c 2 — a 2 = 2bc cos A. 

28. Prove the relations (1) of example 25 directly from a figure. 
C x JO 




Regarding AD and DB as directed lines, we have in each figure 

c =AD + DB = bcosA + acosB. (1) 

Similarly, or by symmetry from (1), we have 
a = b cos C + c cos J5, 
and b = a cos C + c cos .4. 



EXERCISE XXIII 

1. State [31], [32], and [33] in words. 

The area of a triangle is equal to one half the product of any two sides 
into the sine of their included angle. 

The area of a triangle is equal to the square root of the product of its 
half perimeter and its half perimeter minus each side separately. 

The area of a triangle is equal to the square of one side into the sines 
of its two including angles divided by twice the sine of its opposite angle. 

2. Given A = 75°, b = 10, c = 40 ; to find F. 

F = 10 x 40 sin 75°/2 = 200 x 9659 = 193.18. 



92 



PLANE TRIGONOMETRY 



3. Find the areas of the triangles in examples 1-4 in Exercise XXI. 
2 F = ab sin C = ac sin B = bc sin A. 
266, b = 352, 



In example 1, a 
rj _ 73° 

log a = 2.42488 

log 6 = 2.54654 

log sin C = 9.98060 

.-. log(2F) = 4.95202 

.-. F = 89540/2 



10 



44770. 
= 31.2, 



In example 3, a = 960, b = 720, 
C = 25° 40'. 

log a = 2.98227 
log& = 2.85733 
log sin C = 9.63662 - 10 
.-. log (2 F) = 5.47622 

.-. F = 299379/2 = 149689. 

In example 4, a = 886, 6 = 747, 
C = 71° 54'. 

log a = 2.94743 
log 6 = 2.87332 
log sin C = 9.97796 - 10 
.-. \og{2F) = 5.79871 

.-. F = 629086/2 = 314543. 

4. Find the areas of the triangles in examples 1-4 in Exercise XX. 
2 F = be sin A = ac sin B = ab sin C. 



In example 2, b = 91.7, C 
^ = 33° Y 9 // . 

log&= 1.96237 
logc = 1.49415 
log sin A = 9.73750-10 
.-. log(2F) =3.19402 

.-. F= 1563.2/2 = 781.6. 



In example 1, a = 145, b = 178, 
B = 41° 10'. In Exercise XX we 
find C = 106° 24' 24". 

log a = 2. 16137 
log 6 = 2.25042 
log sin C = 9.98194 - 10 
.-. log (2 F) = 4.39373 

.-. F= 24759/2 = 12379. 

In example 2, 6 = 573, c = 394, 
B = 112° 4'. In Exercise XX we 
find A = 28° 20' 48". 

log 6 = 2.75815 

logc = 2.59550 

log sin A = 9.67651 

.-. log(2F)= 5.03016 

.-. F = 107193/2 = 53596. 



In example 3, a = 5.98, b = 3.59, 
A = 63° 50'. In Exercise XX we 
find C = 83° 33' 51". 
loga = 0.77670 
log 6 = 0.55509 
log sin C = 9.99726 - 10 
.-. log(2F)= 1.32905 

.-. F= 21.333/2 = 10.666. 

In example 4, a = 140.5, b = 170.6, 
A = 40°. In Exercise XX we find 
d = 88° 41' 38", C 2 = 11° 18' 22". 
loga = 2.14768 
log 6 = 2.23198 
log sin d = 9.99989 
.-. log (2 Fi) = 4.37955 

.-. F x = 23963.3/2 = 11981.7. 
log (ab) = 4.37966 
log sin C 2 = 9.29237 - 10 
.'. log(2F 2 ) 



3.67203 
F 2 = 4699.2/2 



2349.6. 



EXERCISE XXIII 



93 



Find the areas of the triangles in examples 1-4 in Exercise XXII. 

F = ^s{s-a)(s- b) (s - c). 

56, b = 43, In example 3, a = 61.3, b = 84.7, 

c = 47.6. 



In example 1, a 
c = 49. 



.-. s = 74, s -a = 18, 


.-. s = 96.8, 


s — a = 35.5. 


s-6 = 31, s-c = 25. 


s-b= 12.1, 


s - c = 49.2. 


log s = 1.86923 


logs = 


: 1.98588 


log (s- a) = 1.25527 


log (s-a) = 


: 1.55023 


log(s-6) = 1.49136 


log (s-b) = 


: 1.08279 


log (s-c) = 1.39794 


log (s-c) = 


: 1.69197 


.-. log F= 6.01380/2 


.-. log^ = 


: 6.31087/2 


= 3.00690. 




: 3.15544. 


.-. F = 1016.02. 


.-. F = 


: 1430.3. 


In example 2, a = 8.5, 6 = 9.2, 


In example 4, 


a = 705, b = 


c = 7.8. 


c = 639. 




.-. s = 12.75, s - a = 4.25, 


.-. s = 953, 


s - a- 248, 


s-b= 3.55, s-c =4.95. 


s-b = 391, 


s -c = 314. 


logs= 1.10551 


logs = 


: 2.97909 


log (s- a)= 0.62839 


log (s-a) = 


: 2.39445 


log(s-6)= 0.55023 


log (s - &) = 


: 2.59218 


log (s-c)= 0.69461 


log (s - c) = 


: 2.49693 


.-. \ogF= 2.97874/2 


.-. logF = 


; 10.46265/2 


= 1.48937. 




: 5.23133. 


.-. F = 30.858. 


.-. F = 


: 170346. 



Find the areas of the triangles in examples 1-4 in Exercise XIX. 
2 F = b 2 sin A sin C/sin B = a 2 sin B sin C/sin A. 



In example 1, a = 100, 5 = 60° 15', 
C = 54° 30'. 
A = 180° - (B + C) = 65° 15'. 
log (a 2 ) = 4.00000 
log sin B= 9.93862 -10 
log sin C = 9.91069 - 10 
.-. log product = 13.84931 - 10 
log sin A = 9.95815 - 10 
.-. \og(2F)= 3.89116 
.-. F= 7783.2/2 
= 3891.6. 



45° 4K 



In example 2, b = 100, A 
C = 62° 5'. 
B = 180° - (A + C) = 72° 14' 
log(6 2 )= 4.00000 
log sin A = 9.85460 -10 
log sin G = 9.94627 - 10 
^product = 13.80087 - 10 
log sin B= 9.97878 -10 
log(2F)= 3.82209 
.-. F= 6638.8/2 
= 3319.4. 



l0£ 



94 



PLANE TRIGONOMETRY 



In example 3, a = 102, B = 70° 30', 
C = 78° i0'. 
A = 180° - (B + C) = 31° 20'. 
log (a 2 ) = 2.00860 x 2 
= 4.01720 
log sin 5= 9.97435-10 
log sin C = 9.99067 - 10 
.-. log product = 13.98222 - 10 
log sin A = 9.71602 -10 
.-. log(2F) = 4.26620 
.-. F = 18458.7/2 
= 9229.4. 



In example 4, c 
B = 65°. 



270, A = 55 c 



= 180° -(A + B) = 60°. 
Jogc 2 = 2.43136 x 2 
= 4.86272 
log sin A = 9.91336-10 
log sin B = 9.95728 - 10 
log product = 14.73336 - 10 
log sin C = 9.93753 - 10 
.-. log(2F)= 4.79583 
.-. F = 62493/2 
= 31246. 



EXERCISE XXIY 

1. Express in radians two positive and two negative angles each of 
which is coterminal with ?r/4 ; 5 7T/4 ; 3 7t/2 ; 5 it/2 ; 7T/3 ; 2 7r/3 ; it/Q ; 
5 rt/6. 

■27t + 7t/4:, — 47T + 7T/4 are coterminal with 7r/4 
7r + 7r/4, — 37T + 7T/4 are coterminal with 57T/4 
■7r + 7r/2, — 37T + 7T/2 are coterminal with 3it/2 
2^r + 7T/2, — 47T + 7T/2 are coterminal with 5 7r/2 
27T + 7r/3, — 47T + 7T/3 are coterminal with tt/3 
27T+27r/3, — 47r + 2^r/3 are coterminal with 27T/3 
— 4 7T -j- 7T/6 are coterminal with 7t/6 
— 47T+57T/6 are coterminal with 57T/6. 

6. Express in radians 45°. 



27T+7T/4, 4;r + ^/4, 
3^r + 7r/4, 57T+7T/4, 

37T+7T/2, 57T + 7T/2, 

7T/2, 47T+7T/2, 

27T + 7T/3, 47T+7T/3, 

27r+27r/3,4^r + 27r/3, 



2tt+^/6, 4^r+7T/6, -27T + 7r/6, 
2^ + 57T/6, 4^+ 57T/6, -2^+5^/6, 

2. Express 2 7T/3 in degrees. 
(2/3) 7T = (2/3) 180° = 120° 

3. Express 5 tt/3 in degrees. 

(5/3)7r = (5/3)180° = 300 c 

4. Express 5 it in degrees. 

5 it = 5 • 180° = 900°. 

5. Express 3 7T/4 in degrees. 

(3/4) it = (3/4) 180° = 135 c 

10. Express in radians 97° 25'. 

97° 25' = 97 T 5 3° = (97-^) (*/180) radians 



45° = (1/4) 180° = 7T/4. 

7. Express in radians 135°. 
135° = (3/4) 180° = 3 7t/4. 

8. Express in radians 90°. 
90° = (1/2) 180° = it/2. 

9. Express in radians 270°. 
270° = (3/2) 180°= 3 7t/2. 



1.7 radians. 



EXERCISE XXIV 95 

11. Express in radians 175° 13'. 

175° 13' = (175if) (7T/180) radians 

= 1 W* • ¥ • rio = 3.06 radians. 

12. Express in radians 43° 25' 36". 

43° 25' 36" = (43±£f) - 2 / • T i^ = 0.76 radian. 

13. Express in radians 38° 17' 23". 

38° 17' 23" = (38ftft&)¥ ■ T | 7 = 0.67 radian. 

14. Find the complement and the supplement of it/4. 

comp. = 7T/2 - tt/4 = 7T/4 ; sup. = it — 7T/4 = 3 ar/4. 

15. Eind the complement and the supplement of 2 7T/3. 

comp. = it/2 — 2 tt/3 = — it/6 ; sup. = it — 2 tt/3 = nr/3. 

16. Eind the complement and the supplement of 3 it/4. 

comp. = it/2 — 3 it/4 = — 7r/4 ; sup. = it — 3 7r/4 = 7r/4. 

17. Find the complement and the supplement of 5 it/S. 

comp. = it/2 - 5 7T/3 = - 7 tt/6; sup. = tt - 5 ^r/3 = - 2 tt/3. 

18. Find the trigonometric ratios of it/6. 

it/6 = 180°/6 = 30° ; hence see § 11. 

19. Find the trigonometric ratios of 7T/4. 

7T/4 = 180°/4 = 45°; hence see § 10. 

20. Find the trigonometric ratios of it/S. 

it/S = 18073 = 60°; hence see § 11. 

21. Find the trigonometric ratios of it/2. 

it/2 = 90° ; hence see § 27. 

22. Find the trigonometric ratios of it. 

it = 180° ; hence see § 27. 

23. Two angles of a triangle are 1/2 and 2/5. Find the third angle in 
radians, also in degrees. 

Third angle = 22/7 - 1/2 - 2/5 = 157/70 = 2.24 

= 157(57°17 / 44.8")/70. 

24. The difference between the two acute angles of a right-angled 
triangle is it/1 . Find the angles in radians, also in degrees. 

Let A and B denote the angles required ; then 
(A + B)/2 = 7T/4 = 11/14. \ .-. A = 90/98 = 99(57° 17' 44.8")/98. 
{A - B)/2 = it/14, = 11/49. J B = 55/98 = 55 (57° 17' 44.8")/98. 



96 PLANE TRIGONOMETRY 

25. Express in radians, also in degrees, the angle subtended at the 

center of a circle by an arc whose length is 15 ft., the radius of the circle 

being 20 ft. 

Angle = 15/20 radian = 3/4 radian 

= 3(57°17'44.8")/4. 

26. The diameter of a graduated circle is 6 ft., and the graduations on 
its rim are 5' apart. Find the distance from one graduation to the next. 

Rim = 2 itr in. = (44/7) 36 in. = (1584/7) in. 

Number of graduations = 360 • 12 = 4320. 

Hence the distance required = (1584/7) -=- 4320 = (11/210) in. 

27. Find the radius of a globe which is such that the distance between 
two places on the same meridian whose latitudes differ by 1° 10' may be 
half an inch. 

360°/l° 10' = 308f. 

.-. 2 itr = (308f) (1/2) in. = 154f in. 
.-. r = (77£) (7/22) in. = 24 T B T in. 

28. The value of the divisions on the outer rim of a graduated circle 
is 5' and the distance between successive graduations is 0.1 in. Find 
the radius of the circle. 

36075' = 360 • 12. 
.-. 2 itr = 360 -12-0.1 in. = 432 in. 
.-. r = 216 ■ 7/22 in. = 68 r 8 T in. 

29. Assuming the earth to be a sphere and the distance between two 
points 1° apart to be 69| mi., find the radius of the earth. 

Circumference = 360 • 691 m i. - 25020 mi. 

.-. 2 itr = 25020, or r = 25020(7/44) = 3980 T 5 T . 



EXERCISE XXV 

1. Solve sin 2 = l. 

sin = ± 1 = sin (± it/2). 
.-. = n7t + (- l) n {± it/2) = U7C± 7t/2. 

2. Solve esc 2 = 2. 

Solution 1. esc = ± V2 = esc (± it/4). 

.-. 6 = nit + (- l) n (± 7t/4) = mt ± 7t/4. 
Solution 2. sin 2 - 1/2, or sin = ± 1A/2. 

.-. = nit ± it/i. 



EXERCISE XXV 



97 



3. Solve tan 2 



4. Solve cot 2 1 



tan = ± 1 = tan(± #/4). 

.-. = mt + (± 7T/4) = Mt ± 7T/4. 

cot = ± V3 = cot(± 7t/Q). 
.-. ^ = n?r + (± 7T/6) = n?r ± zr/6. 



5. Solve cos 2 = 


= 1/4. 


cos = ± 1/2. 


Let 




cos0i = 1/2 = cos(7r/3), 


and 




cos 6 2 = — 1/2 = cos (4 7T/3). 


Then 




0i = 2 n X Tt ± 7T/3, 


and 




2 = 2 ?i 2 7r ± 4 tf/3 = (2 ft 2 ± 1) 7T ± 7T/3 
.-. = ?17T ± It/Z. 


6. Solve sec 2 B- 


:4/3. 




Solution 1. 




sec = ± 2/V3. 


Let 




sec 0i = J- 2/^/3 = sec (tt/6), 


and 




sec 2 = - 2/V3 = sec (5 ar/6). 
.-. 6-1 = 2 n x it ± 7T/6, 


and 




2 = 2 n 2 7t ± 5 7T/6 = (2 n 2 ± 1) it T tt/6 
.'. = 717T ± 7T/6. 


Solution 2. 




cos 2 = 3/4. 
. cos = ± V3/2, etc. 


7. Solve 2sin 2 + 3 cos 


= 0. 


By [2], 2-2 


cos 2 + 3cos0 = 0. 


.-. (2cos0 


+ 1) (cos 


- 2) = 0. 



cos = 2 is impossible, 
cos = - 1/2 = cos (2 7T/3). 
.-. = 2 m ± 2?r/3. 

8. Solve cos 2 — sin = 1/4. 
By [2] , 1 - sin 2 - sin = 1/4. 

.-. sin 2 + sin0-3/4 = 0. 
.-. (sin + 3/2) (sin - 1/2) = 0. 

sin = — 3/2 is impossible. 
sin0 = 1/2 = sin (n/Q). 
.-. = mt + (-l)"7T/6. 



98 PLANE TRIGONOMETRY 

9. Solve 2 V3 cos 2 5 = sin 5. 
By [4], 2 V3 - 2 V3 sin 2 5 = sin 5. 

.-. 2 V3 sin 2 5 + sin 5 - 2 V3 = 0. 
.-. (V3 sin 5 + 2) (2 sin 5 - V3) = 0. 

sin = — 2/ V3 is impossible, 
sin = V3/2 = sin (it/ 2). 
.-. 6= ?wt + (-l)»ar/3. 

10. Solve sin 2 5 - 2 cos 6 + 1/4 = 0. 

By [4], 1 - cos 2 5 - 2 cos + 1/4 = 0. 

.-. cos 2 5 + 2 cosfl - 5/4 = 0. 

.-. (cos + 5/2) (cos - 1/2) = 0. 

cos# = — 5/2 is impossible, 
cos 6 = 1/2 = cos (it/3). 
.-. = 2nit ± it/2,. 

11. Solve sin + cos = V2. (1) 
By (1), sin0 and cos0 must both be positive; hence is in the first 

quadrant. 

Square (1), sin 2 + cos 2 + 2 sin cos = 2. (2) 

By [4], from (2), 2 sin cos 5 = 1. 

By [13], sin2 = l. 

Hence 2 5 is coterminal with it/2. 

.-. 2 5=2 W7T + it/2, or = n?t + it/4:. (3) 

But as is in the first quadrant, 5 = 2 n7T + it/ 4. 

The other solutions in (3) were introduced by squaring (1). 

12. Solve 4 sec 2 0-7 tan 2 5 = 3. 

By [5], 4 tan 2 5 + 4-7 tan 2 5 = 3. 

.-. 3 tan 2 5 = 1. 
.-. tan 5 = ± 1/V3 = tan (± ar/6). 
.-. 5 = n7T ± it/6. 

13. Solve tan 5 + cot 5 = 2. 

.-. tan 5 + l/tan5 = 2. 

.-. tan 2 5 - 2 tan 5 + 1 = 0. 

.-. (tan 5 - l) 2 = 0. 

.-. tan 5 = 1 = tan (it/4.). 
.-. = nit + it/4:. 

14. Solve tan 2 5 - (1 + y/S) tan 5 + V3 = 0. 
Factor, (tan 5-1) (tan 5 - V3) = 0. 

.-. tan 5 = 1 = tan (it/4), or tan 5 = V3 = tan (it/ 3). 
.-. 5 = nit + 7T/4, or 5 = nit + 7T/3. 



EXERCISE XXV 99 



15. 


Solve cot 2 + (V3 + W3)cot0 + 1 = 0. 
(cot + V3) (cot + 1/V3) = 0. 
cot = - V3 = cot (5 7T/6), or cot = - 1/V3 = cot 
.-. = n7T + 5 7T/6, or = ri7r + 2 7r/3. 


(2 


7T/3). 




16. 


Solve tan 2 + cot 2 = 2. 

.-. tan 2 + l/tan 2 = 2. 

.-. tan 4 0- 2tan 2 + l = 0. 

.-. (tan 2 0-l) 2 = 0. 

.-. tan0 = ± 1 = tan(± 
.-. = mt ± 7t/4. 


*/4). 








17. 


Solve tan + sec = 3. 


7' 45". 

7' 45". 






(1) 


.-. tan + Vl + tan 2 = 3. 

.-. Vl + tan' 2 = 3 - tan 0. 
Squaring, G tan — 8 = 0. 

tan = 4/3 = tan 53 c 
6 = n • 180° + 53 c 




But 


(1) is not satisfied when n is odd, that 


is, when 


is 


in the third 



18. Solve 2 sin = 1 + cos 0. (1) 

.-. 2 sin - 1 = Vl - sin 2 0. 
.-. 5sin 2 - 4 sin = 0, or sin 0(5 sin - 4) = 0. 

.-. sin = = sin 0, or sin = 0.8 = sin 53° 8 X . 
.-. = ?i-18O o , or0 = n-18O°+(-l)'*53 o 8 / . 
But (1) is not satisfied when n in n ■ 180° is even, or when n in n ■ 180° + 
53° 8' is odd ; hence = (2 n + 1) 180°, or 2 n ■ 180° + 53° 8'. 

19. Solve sin 5 = 1/V2 = sin (tt/4). 

.-. 50 = nit + (-l)*7r/4. 
.-. = [n7r + (-l)"7r/4]/5. 

20. Solve cos 5 = cos 4 0. 

.-. 5 = 2 nit ± 4 0. 
.-. = 2 nit, or 2 n7r/9. 

21. Solve cot = tan rd. 

.-.- tan r-0 = cot = tan (it/ 2 - 6) . 

.-. r9 = nit + 7T/2 - 0. 
.-. d = (nit+it/2)/(r+l). 



100 PLANE TRIGONOMETRY 

22. Solve sin 2 = cos 3 0. 

sin 2 = cos {it/ 2 - 2 0). 
.: cos 3 = cos (it/2 -2 0). 
.: S0 = 2n7t + ar/2 -2 0, 
or 3 = 2mt - 7T/2.+ 2 0. 

.-. = (2n7T + tt/2)/5, 
or = 2 fttf - ar/2. 

23. Solve cos m,0 = sin rO. 

sin r0 = cos {n/2 — rO). 
.-. cosra0 = cos(7T/2 — rO). 
.: mO - 2 nit ± -rt/2 =F r0. 
.-. = (2mt ± 7t/2)/(m ± r). 

24. Solve sin cos = 1/2. 

2 sin cos = sin 2 0. 
.-. sin2 = 1. 
Since sin 2 = 1, 2 is coterminal with n/2. 

.-. 2 0= lt/2 + 2 ?17T. 

.-. = mt + 7t/4. 
must be in the first or third quadrant, because sin cos is positive. 

25. Solve sin cos = - V3/4. 

2 sin cos = sin 2 0. 

.-. sin 2 = - V3/2 = sin ( - ?r/3) . 

.•; 20 = n7T + (-l) w (- 7T/3). 

If n is even, or if n = 2 m, 2 0=2 ni7T — 7T/3. 

.-. = m7T — 7T/6. 

If n is odd, or if n = 2»i + l, 2 = (2m + l)7r + 7T/3. 

.'. = Witf + 7T/2 + 7T/6 = 7117T + 2 7T/3. 

26. Solve sec csc = - 2. 

sec esc = 2/(2 sin cos 0) = 2/sin 2 0. 
.-. sin 2 = — 1. 

Since sin 2 = — 1, 2 is coterminal with — it/2. 

.-. 2 = - 7T/2 + 2 ntf, or = mt - it/±. 
must be in the second or fourth quadrant, since sec esc is negative. 

27. Solve tan 2 tan = 1. 

.-. tan 2 = 1/tan 0. 

1/tan = cot = tan {it/2 - 0). 
.-. tan 2 = tan(?r/2 - 0). 
' .-. 20 = n?r + {tc/2 - 0). 
.-. = (nit + tt/2)/3. 



EXERCISE XXV 101 

fc 28. Find the most general value of 6 that satisfies cos0 = — 1/V2 and 
tan0 = 1. 

Since cos 6 is — and tan 6 is + , 6 is in the third quadrant. 
.-. cos 6 = - 1/V2 = cos (5 7T/4). 
tan 6 = 1 = tan (5 7r/4). 

Hence the angles coterminal with 5 7T/4, and only these, satisfy the 
conditions. ... d _ 2 n7r + 5 ?r/4 = (2 n + 1) tt + ff/4. 

29. Find the most general value of that satisfies cot 8 = — V3 and 
esc 6 = - 2. 

Since cot is — and esc 6 is — , 6 is in the fourth quadrant. 

Hence cot 6 = — V<3 = cot ( — it/Q). 

csc0 = — 2 = csc(— 7T/6). 

.'. = 2 ?17T - 7C/6. 

30. If cos (J. - B) = 1/2 and sin (^4 + B) = 1/2, find the smallest posi- 
tive values of A and B, and also their general values. 

cos (^4 -B) = 1/2 = COS(7T/3). 

sin (4+ B) = 1/2 = sin (tt/6), or sin(5 7r/6). 

Hence to obtain the smallest positive values of A and B, we have 
A - B = tt/3, A + B = 5 ?r/6. 
.-. ^4 = 7 7T/12, B = 7T/4. 

To obtain the general values of A and B, we have 

J. - B = 2 ?W7T ± 7T/3, 

J. + .B = ft7zr + (- 1)"tt/6. 

.-. ^4 = (2 in + n) tt/2 ± x/Q + (- l) n ?r/12. 
5 = (n -2m)7t/2 =F tt/6 + (- 1)»tt/12. 

31. If tan (A- B) = l and sec (J. + B) = 2/V3, find the smallest posi- 
tive values of A and 5, and also their general values. 

tan (A - B) = 1 = tan (7T/4). 

cos (4 + B) = V3/2 = cos (tt/6), or cos (11 it/6). 

For the smallest positive values, we have 

A - B = tt/4, .4 + J3 = 11 tt/6. 
.-. ^4 = 25 tt/24, .8 = 19 tt/24. 

For the general values, we have 

(A - B) = mrc + tt/4, ^4 + 5 = 2 n?r ± tt/6. 

.-. A = (2 71 + m) 7T/2 + 7T/8 ± 7T/12. 

5 = (2n- m) tt/2 - ?r/8 ± tt/12. 



102 PLANE TRIGONOMETRY 

EXERCISE XXVI 

1. Solve sin + sin 7 = sin 4 0. 

sin + sin 7 = 2 sin 4 cos 3 0. 
.-. 2 sin 4 cos 3 = sin 4 0, 
or sin 4 0(2 cos 3 - 1) = 0. 

From sin 4 = = sin 0, 

4 = mt. 
From cos 3 = 1/2 = cos (tt/3), 

Sd = 2me± 7T/3. 
.-. = n7r/4, or (2 mt ± tt/3)/3. 

2. Solve cos + cos 3 = cos 2 0. 

cos + cos 3 0=2 cos 2 cos 0. 

.-. 2 cos 2 cos = cos 2 0, 
or cos 2 (2 cos - 1) = 0. 

From cos 2 = = cos (7T/2), 

2 = 2 ntf ± 7T/2, or = n7r ± tt/4. 
From cos0 = 1/2 = cos(± tt/3), 

= 2 mt ± 7t/S. 

3. Solve sin — sin 3 = cos 2 0. 

sin - sin 3 = - 2 cos 2 sin 0. 
.-. — 2 cos 2 sin = cos 2 0, 
or cos 2 0(2 sin + 1) = 0. 

From cos 2 = = cos (?r/2), 

2 = 2 ntf ± 7T/2, or = W7T ± nr/4. 
From sin = - 1/2 = sin (- 7T/6), 

0= >i7r + (-l) M (-7r/6) = 7i7r-(-l)".^/6. 

4. Solve cos + cos 7 = cos 4 0. 

cos + cos 7 = 2 cos 4 cos 3 0. 
.-. 2 cos 4 cos 3 = cos 4 0, 
or cos 4 0(2 cos 3 - 1) = 0. 

From cos 4 = = cos (ar/2), 

4 = 2 n7T ± 7T/2, or = n;r/2 ± tf/8. 
From cos 3 = 1/2 = cos (7T/3), 

3 = 2 ntf ± 7T/3, or = 2 mt/% ± it/9. 

5. Solve sin 4 - sin 2 = cos 3 0. 

sin 4 - sin 2 = 2 cos 3 sin 0. 
.-. 2 cos 3 sin = cos 3 0, 
or cos 3 (2 sin - 1) = 0. 



EXERCISE XXVI 103 

From cos 3 = = cos (it/ 2), 

30 = 2n7t ± 7T/2, or = 2 n;r/3 ± tt/6. 
From sin0 = 1/2 = sin (7t/6), 

= 717* + (- l)»7T/6. 

6. Solve sin 7 = sin + sin 3 0. 







.-. sin 


re 


— sin = sin 3 0. 






sin 


id 


- sin = 2 cos 4 sin 3 0. 






•. 2 cos 4 sin 3 = sin 3 0, 


or 


sin 


3 0(2 cos 4 -1) = 0. 


From 








sin 3 = = sin 0, 

3 = wt. 


From 








cos 40 = 1/2 = cos(tt/3), 

4 = 2ii7t ± 7T/3 = (2n± 1/3) 7f 
.-. = 7i7r/3, or (2 w ± 1/3) tt/4. 



7. Solve sin 0+ sin 3 + sin 5 = 0. 

sin 5 + sin = 2 sin 3 cos 2 0. 
.-. sin 3 + 2 sin 3 cos 2 = 0, 
or sin 3 0(2 cos 2 0-f 1) = 0. 

From sin 3 = = sin 0, 

= nit/Z. 
From cos 20=- 1/2 = cos(2tt/3), 

-mt ± 7t/S. 

8. Solve cos + cos 2 + cos 3 = 0. 

cos 3 + cos = 2 cos 2 cos 0. 
.-. 2 cos 2 cos + cos 2 = 0, 
or cos 2 (2 cos + 1) = 0. 

From cos 2 = = cos (tt/2) , 

= nit ± tt/4. 
From cos0 = - 1/2 = cos(2 7T/3), 

= 2 J17T ± 2 7T/3. 

9. Solve V3 cos + sin = V2. (1 ) 
Solution 1. Dividing by V( V3) 2 + l 2 , or 2 (§ 73), we obtain 

( V3/2) cos + (1/2) sin = V2/2, (2) 

V3/2 = cos (tt/6), 
1/2 = sin (tt/6), 
V2/2 = cos(tt/4). 



104 PLANE TRIGONOMETRY 

Substituting these values in (2), we have 

cos (it/6) cos + sin (it/6) sin = cos (nr/4), 
or cos(0 — it/6) = cos(7r/4). 

.-. - it/6 = 2 nit ± tt/4, 
or e = 2 nit + 7T/6 ± tf/4. 

Solution 2. Here ^4 = tan- 1 (1/ V3) = tt/6, 5 = cos- 1 ( V2/2) = ar/4. 
.-. 6 = 2nit + it/6 ± 7T/4. 

10. Solve sin + cos = y/2. 

Here A = tan- 1 1 = it/ 4, B = cos- 1 1 = 0°. 
.-. 6 = 2 nit + it/4. 

11. Solve V3 sin 8 - cos 6 = -y/2. 

Making the coefficient of cos positive, we obtain 

- V3 sin + cos = - V2. 

Here ^ = tan- 1 - V3 = - tf/3, 

and J5=cos-^- — ) = 3tt/4. 

.-. 0=2?i7r- 7T/3 ± 3tt/4. 

Since A is the principal value of tan _1 (<x/6), cos A is positive, hence b 
must be positive in § 73. 

12. Solve sin + cos = V 2 cos (^ /§)• 

Here A = tan- x 1 = it/4, 

and 5 = cos -1 cos(7r/5) = it/ 5. 

.-. = 2 nit + 7T/4 ± it/b. 

13. Solve 5 sin + 2 cos = V29, given tan 68° 12' = 5/2. 

Here A = tan- 1 5/2 = 68° 12', 

and jB = cos - T 1 = 0. 

= 2^180° + 68° 12'. 

14. Solve 2 sin - 3 cos = VW2, given tan 33° 41' 24" = 2/3. 
Making the coefficient of cos positive, we obtain 

- 2 sin + 3 cos = - V13/2. 

Here ^4 = tan- 1 (- 2/3) = - 33° 41' 24", 

and J5 = cos- 1 ( - 1/2) = 120°. 

.-. = 2 n 180° - 33°41 / 24 // ± 120°. 



EXERCISE XXVII 105 

EXERCISE XXVII 

1. Find the value of vers (#/6), vers (tt/4), vers (7T/3), vers (3 7T/4), 
vers 0, vers (?r/2), vers tt, vers (3 7t/2). 

Since vers x = 1 — cos x, 

vers (7t/6) = 1 - sJZ/2 = (2 - V3)/2, 

vers (tt/4) = 1 - V2/2 = (2 - V2)/2, 

vers (tt/3) = 1 - 1/2 = 1/2, 

vers (3 tt/4) = 1 - (- V2/2) = (2 + V2)/2, 

vers =1-1=0, 

vers (7t/2) =1-0 = 1, 

vers tc =1 — (— 1) = 2, 

vers (3 it/2) = 1-0 = 1. 

2. Find the value of covers (ar/6), covers (7r/4), covers (7T/3), covers 
(3 7T/4), covers 0, covers (tf/2), covers 7T, covers (3 7T/2). 

Since covers x = 1 — sin x, 

covers (tt/6) = 1 - 1/2 = 1/2, 
covers {Tt/4) = 1 - V2/2 = (2 - V2)/2, 
covers (tt/3) = 1 - V3/2 = (2 - V3)/2, 
covers (3 tt/4) = 1 - V2/2 = (2 - V2)/2, 
covers (0) =1-0 = 1, 
covers (it/2) =1-1 = 0, 
covers (it) =1—0 = 1, 
covers (3 tt/2) = 1 - ( - 1) = 2. 

3. Express in radians the general value of 

sin-i(V2/2); sin- 1 ( - V3/2) ; cos-i(V3/2); cos-* (-1/2); 
tan-i(V3/3); tan-!(-V3); cot- 1 (-l); cot-i(V3/3). 

V2/2 = sin (tt/4); .-. sin-i(V2/2) =nit + (- l)«tf/4. §69 

- V3/2 = sin ( - Tt/3) ; .-. sin- 1 ( - V3/2) = mt - (- 1)» tt/3 . 

V3/2 = cos(^r/6); .-. cos" 1 ^^) = 2n7r±7r/6. §70 

- 1/2 = cos (2 Tt/2) ; .\ cos- * ( - 1/2) =2n7t ±2 tt/3. 

V3/3 = tan (tt/6); .-. tan-i(V3/3) = nit + Tt/6. § 71 

- V3 = tan ( - Tt/3) ; .-. tan- x ( - V3) = »* - tt/3. 
- 1 = cot(- tt/4); .-. cot- 1 (-l) =riTt-Tt/4. 

V3/3 = cot (tt/3); .-. cot" 1 ( V3/3) = mt + #/3. 



106 PLANE TRIGONOMETRY 

4. Prove tan- 1 2 + tan- x 0. 5 = tt/2. (1) 

Let 4= tan- 1 2, £ = tan- 1 (0.5). (2) 

Then tan 4 =2, tan B = 0.5. (3) 

From (1), (2) 4 + J? = ar/2, or tan (J. + 5) = tan (tit/2) = oo. 

From (3), tan (4 + B) = 2 + °- 5 = ». 

W V ' 1-2x0.5 

5. Given 0.5 = tan 26° 33' 53. 5", prove 

tan-!7 +tan-i3= 153°26 / 6.5 // . (1) 

Let 4 = tan- 1 7, £ = tan-!3. (2) 

Then tan 4 = 7, tan B = 3. ■ (3) 

From (1), (2) A + B = 153° 26' 6.5", or tan (A + B) = - 0.5. 

By (3), tan (A + B)= 7 + 8 = - 0.5. 

1 — o • 7 

6. Prove tan -1 tan -1 = - • (1) 

n - m + n 4 

Let 4 = tan- 1 -, Bee tan- 1 — — . (2) 

n m + n 

Then tan^i ee-, tan £ = 7?1 UZ^' . ( 3 ) 

n m + n 

From (1), (2), 4 - £ =. tt/4 ; .-. tan (4 - B) = tan (tt/4) = 1. 
m m — n 



-r. /ov / „ ™ w m + n _ m 2 + n 2 

By (3), tan (4 - B) = = — = 1. 

m m - n m 2 + w 2 

n ?n + ?i 

7. Prove tan- 1 (1/7) + tan- 1 (1/13) = tan- 1 (2/9). (1) 

Let 4 = tan- 1 (1/7), B = tan- 1 (1/13). (2) 

Then tan A = 1/7, tan B = 1/13. (3) 

From (1), (2), A + B = tan- 1 2/9, or tan (A + B) = 2/9. 

1/7 + 1/13 _ 20/91 _ 2 



From (3), tan (A + B) 



1 - 1/91 90/91 9 



8. Prove 2 tan- * (2/3) = tan- 1 (2/3) + tan - 1 (2/3) = tan- * (12/5). (1) 

Let 4 = tan- 1 (2/3). 

Then tan A = 2/3. (2) 

From (1), (2), 2 A = tan- 1 (12/5), or tan (24) = 12/5. 
2 • 2/3 12/9 12 



From (2), tan 2 A 



1 - 4/9 5/9 



EXERCISE XXVII 107 



9. tan- 


i (3/4) + tan- * (3/5) - tan" 1 (8/19) = tt/4. 


(1) 


Let 


A = tan- 1 3/4, B = tan- 1 (3/5). 


(2) 


Then 


tan A = 3/4, tan B = 3/5. 




.-. ton (A + B)~ 8/4 + 8/5 = 27 , or A + B = tan- 1 (27/11). 
v ' 1 - (3/4) (3/5) 11 


(3) 


Let 


C = ton- 1 (27/11), D = tan- 1 (8/19). 


(4) 


Then 


tan C = 27/11, tan D = 8/19. 




From (1), 


(2), (3), (4), C -D= Tt/i, or tan (C - D) = tan (tt/4) = 


1. 


From (4), 


tau(C D) = 27/11 - 8/10 = l. 
x 27X8 

11 x 19 




10. Prove sin- 1 (3/5) + sin- 1 (8/17) = sin- 1 (77/85). 


(1) 


Let 


A = sin- 1 (3/5), B = sin- 1 (8/17). 


(2) 


Then 


sin A = 3/5, sin 5 = 8/17. ^ 
cos A = 4/5, cos B = 15/17. J 


(3) 


Hence 


From (1), 


(2), A + B = sin- 1 (77/85), or sin (4 + JB) = 77/85. 




By (3), 


sin (A + B) = sin 4 cos 5 + cos A sin 2> 

= 3.154.4, _8 = 7 7 
— 5 17 T 5 IT — 85- 





11. Prove cos- 1 (4/5) -Kcos- 1 (12/13) = cos- 1 (33/65). (1) 
Let A = cos- 1 (4/5), B = cos- 1 (12/13). 

Then cos A = 4/5, cos B = 12/13. 1 

.-. sin A = 3/5, sin B = 5/13. J 

Hence 4 + .B = cos- 1 (33/65), or cos (4 + 5) = 33/65. 

By (2), cos (A + B) =~$ . if - f . £ = ff . 

12. Prove cos- ! (4/5) + tan- 1 (3/5) = tan- 1 (27/11). 
Let 4 = cos- 1 (4/5), B = tan- 1 (3/5). 

Then cos A = 4/5, tan 4 = 3/4, tan 72 = 3/5. (1) 

.-. 4 + B = tan- 1(27/11), or tan (A +B) = 27/11. 

By(l), tan(4 + B) = 3/4 + 8/5 = g. 
W ' V ' 1- (3/4) (3/5) 11 

13. Solve ton~ l x + tan- 1 (1 - x) = tan- 1 (4/3). 
By [36], tan- 1 x + tan- 1 (1 - x) = tan- 1 



1 - x + x' 2 

tan- » = tan- 1 4/3. 

1 - x + x 2 

1 4 



1 - x + x 2 3 
... 4x 2 -4x + l = 0. 
.-. (2x-l) 2 = 0. 
.-. a; = 1/2. 



108 PLANE TRIGONOMETRY 



_.', , .x — 1 ,x + 1 it 

14. Solve tan- 1 \- tan- 1 — — = -. 

x - 2 x + 2 4 

By [36], tan-i^^i + tan-i£±i = tan~ l 4 ~ 2a;2 . 

LJ x-2 x + 2 3 

4 - 2 x 2 ?r n 

.-. = tan - = 1. 

3 4 

.-. x = ± VI72 = ± V2/2. 

15. Solve tan- 1 & + 2cotr 1 x = 2 7r/3. (1) 

Let A = t£iJi- 1 x, B = cot - l x. 

Then tan A = x, cot J5 = x, tan 2? = 1/x. (2) 

. _._ 2tan£ 2x _ w . . 2x ... 

tan 2 5 = = , or 2 i? = tan- L (3) 

1 - tan 2 5 x 2 -1 x 2 -l w 

.-. i + 2B = tan-ix + tan- 1 -^- = tan- 1 a 



x* + x 2* X2 ~ 1 -<* 2 + 1 > 

.*. tan- 1 ; = 

- (x 2 + 1) 3 

x 3 + x 2tt . 

:-. = — tan — = V 3 - 

x 2 + 1 3 

■. x 3 - V3 x 2 + x - V3 = 0, or (x - V3) (x 2 + 1) = 0. 

.-. x = V3, ± V^T ; of which only V 3 is admissible. 

16. Solve tan-i (x -f 1) + tan- 1 (x - 1) = tan- 1 8/31. 

2 x 
tan- 1 (x + 1) + tan- 1 (x — 1) = tan- 1 • 

Hence 2x/(2 - x 2 ) = 8/31. 

.-. 4x 2 + 31»-8 = 0. 
.-. (x + 8)(4x- 1)=0. 

.-. x=- 8, 1/4. 

,x 2 -l . 2 x 2tt 

17. Solve cos- l h tan- 1 — = -— . 

x 2 + 1 x 2 - 1 3 

Let J. = cos- 1 [(x 2 - l)/(x 2 + 1)], B = tan- 1 [2 x/(x 2 - 1)]. 

Then cos A = (x 2 - l)/(x 2 + 1), tan £ = 2 x/(x 2 - 1). 

.-. sin A = 2 x/(x 2 + 1), tan A = 2 x/(x 2 - 1). 

.-. tan A = tan 5, or A = B. 

.-. tan-i[2x/(x 2 -l)] = ar/3. 

.-. 2 x/(x 2 - 1) = tan (tt/3) = ^/3. 

.-. x 2 - 2 V3 x + 3 = 0, or (x - V3) 2 = 0. 

.-. x = V3. 



EXERCISE XXVIII 109 

18. Solve sin-ix + sin- 1 2 z = tt/3. (1) 
Let A^sin-^x, B = sm- 1 2x. (2) 
Then sin A = x, sin B = 2 x, 

.-. cosJ. = Vl -x 2 , cos£eeV1-4x 2 . 

From (1), by (2), 

A + B = 7t/il, or cos (A + B) = cos (tt/ 3) = 1/ 2. (3) 

cos ( A + B) = cos A c os B - si n A sin B = Vl - x 2 Vl - 4 x 2 - 2 x 2 . (4) 
From (3), by (4), Vl - x 2 Vl-4x 2 - 2 x 2 = 1/2. 

.-. 7 x 2 = 3/4, or x = ± -y/21/U. 

19. Solve sin- 1 (5/x) + sin- 1 (12/x) = n/2. (1) 
Let J. = sin- 1 (5/x), B = sin- 1 (12/x). (2) 
Then sin J. = 5/x, sin 5 = 12/x, cos A = 12/x, cos B = 5/x. 

From (1), by (2), 

A + B = 7t/2, or sin (A + B) = sin (tt/2) = 1. (3) 

sin (A + B) = sin ^L cos B + cos vl sin B = (5/x) 2 + (12/x) 2 = 169/x 2 . (4) 
From (3), by (4), x 2 = 169, or x = ± 13. 



EXERCISE XXVIII 

1. Prove the identity tanh 2 x + sech 2 x = 1. 

Solution 1. tanh 2 x + sech 2 x = (e x - e~ x Y/(e x + e~ x ) 2 + 4/(e T + e-^) 2 

= {e x + e~ x ) 2 /(e x + e~ x f = 1. 
Solution 2. By § 86, 

cosh 2 x — sinh 2 x = 1. 
Divide by cosh 2 x, 

1 — sinh 2 x/cosh 2 x = 1/cosh 2 x. 
Hence, by § 85, 1 — tanh 2 x = sech 2 x, or tanh 2 x + sech 2 x = 1. 

2. Prove the identity coth 2 x — csch 2 x = 1. 
By § 86, cosh 2 x - sinh 2 x = 1. 

.-. cosh 2 x/sinh 2 x — 1 = 1/sinh 2 x. 

.-. coth 2 x — 1 = csch 2 x, or coth 2 x — csch 2 x = 1. 

_ _, , . x tanhx + tanhv 

3. Prove tanh (x + y) = — — ■ (1) 

1 + tanh x • tanh y 

By § 86, sinh (x + y) = sinh x cosh y + cosh x sinh y, (2) 

and cosh (x -f y) = cosh x cosh y + sinh x sinh y. (3) 

-rv -a , \ u /o\ . i / , \ sinh x cosh y + cosh a; sinh y 

Divide (2) by (3), tanh (x + y) = • 

cosh x cosh y + sinh x sinh y 

Dividing numerator and denominator by cosh x cosh ?/, by § 85, we 
obtain (1). 



110 PLANE TRIGONOMETRY 

4. Prove sinh 3 x = 3 sinh x + 4 sinh 3 x. (1) 

By §85, sinh3x = (e 3 *- e~ 3a: )/2, (2) 

and 3 sinh x + 4 sinh 3 x = f (e* - e~ a ) + | (e x - e~ x Y 

~ (e 3 * - e-3 x )/2. (3) 

From identities (2) and (3) we have identity (1). 

5. Prove cosh 3 x = 4 cosh 3 x — 3 cosh x. (1) 

By § 85, cosh 3 x = (e" x + e- 3 *)/2, (2) 

and 4 cosh 3 x - 3 coshx = J (e r + e-- T ) 3 - § (e x + e~ x ) 

== (eSo: + e-s^/2. (3) 

From identities (2) and (3) we obtain (1). 



6. Prove sinh- 1 x ee cosh- 1 Vl + x 2 = tanh- 1 (x/vl + x 2 ). (1) 

Let A ee sinh-^x ; then sinh A = x. 



cosh A = Vl -f x 2 , or ^L = cosh- 1 Vl + x 2 . 



Let 


A 


Then 


tanh A 


and 


A + B 


But tanh 


(A + B) 



Also, tanh A = x'/Vl + x 2 , or A = tanh- 1 (x/Vl + x' 2 ). 

As each member of (1) is identical with A, (1) is an identity. 

x -f- v 

7. Prove tanlr^x + tanh- 1 ?/ = tanh- 1 (1) 

1 + xy 

tanh- 1 x, Bee tanh- 1 ?/. (2) 

x, tanh B ee y, (3) 

tanh- 1 ^±^ ■ .-. tanh (A + £) = ^-^1 . 
1 + xy 1 + xy 

tanh ^4 + tanh B _ x -f y 
1 + tanh ^4 tanh B ~ 1 + xy 



EXERCISE XXIX 

Represent graphically and reduce to the trigonometric type form each 
of the following twelve complex numbers : 

1. 1 + V^l. 

Draw OA = + 1 and AP = i ; then OP will represent 1 + V— 1. 
The type form of the complex number a + ib is (a/r + ib/r ) ■ r where a 
and b are real and r denotes the arithmetic value of Va 2 + 6 2 . 

Here a = l, 6 = 1; .-. r =^/2. 

.-. 1 + V^TT = (1/^/2 + i/V2) • V2 = [cos (tt/4) + i sin (tt/4)] • ^2. 



EXERCISE XXIX 111 



2. 1-V-l. 

Draw OA =-+ 1 and AP = — i ; then OP will represent 1 — V— 1. 

Here a = 1, & = — 1; .-. r = V2. 

••• l-VTT = [i/ v / 2 + i(- 1/V2)]- V2 

= [cos(- ?r/4) + isin(- tt/4)] • V2. 

3. -1-2. 

Draw OA = — 1, and J.P = — 2 ; then OP will represent — 1 — i. 
Here a = - 1, b = - 1 ; .-. r = V2. 

.-. - 1 - i = [_ 1/V2 + *(- 1/V2)] • V2 

= [cos(- 3 tt/4) + i sin(- 3 tt/4)] • y/% 

4. _i + V-3. 

Draw OA — — 1 and J.P = 2 V3 ; then OP will represent — 1 + V— 3. 
Here g = -l, 5 = V3; .-. r = 2. 

.-. -1 + V_8= [-l/2 + iV3/2] •2=[cos(2 7r/3) + isin(2 7r/3)] -2. 

5. - V3 + i. 

Draw 0^4 = — V3 and AP = i ; then OP will represent — V3 + i- 
Here a = - V3, b = 1 ; .*• r = 2. 

.-. - V3 + i = ( - V3/2 + 2/2) • 2 = [cos (5 n/Q) + i sin (5 ?r/6)] • 2. 

6. - V3 - t. 

Draw OJ. = — V3 and J.P = — i ; then OP will represent — V3 — i. 
Here a = V3, & = - 1 ; .-. r = 2. 

.-. -V3-i = [-V3/2+i(-l/2>]..2 

= [cos( - 5 tt/6) + i sin ( - 5 ^/6)] • 2. 

7. 3-^4. 

Draw 0^4 = 3 and AP = — i 4 ; then OP will represent 3 — i 4. 
Here a = 3, b = — 4 ; .-. r = 5. 

.-. 3--i4 = [3/5 + i(-4/5)]-5 

= [cos (- 53° 8 r ) + i sin ( - 53° 8')] • 5. 

8. -3 + 22. 

Draw OA = -3, ^4P = 22; then OP will represent -3 + 22. 
Here a = - 3, b = 2 ; .-. r = V13. 

.-. - 3 + 2 2 = (- 3/V13 + 2 2/V13) • V13 

= (cos 146° 19' + 2 sin 146° 19') • V13. 

9. - 6 - 2 8. 

Draw OA = -6, AP = -i8; then OP will represent - 6 - i 8. 
Here a = - 6, & = - 8 ; .-. r = 10. 

.-. _ 6 - 2 8 = [- 3/5 + i(- 4/5)] ■ 10 

= [cos(- 126° 52') + 2sin(- 126° 52')] : 10. 



112 



PLANE TRIGONOMETRY 



10. -V5- i Vll. 

DrawOJ.= — V5and^4P= — i-^11; then OP will represent — -y/5 — iVH- 
Here a = — V5, & = — VHj •'• »" = 4. 

.-. - V5-iVH = [-V5/4 + i(-VH/4)]-4 

= [cos (- 123° 59') + i sin (- 123° 590] ■ 4 - 

11. sin + i cos = cos (7T/2 — 0) + i sin (7t/2 — 0). 

12. — sin 6 + i cos 6 ee cos (?r/2 + 6) + i sin (tt/2 + &). 

13. By constructing the sum in each member, show that 

(a + ib) + (x + i?y) = (a + %) + (a; + ib) = (a + x) + i (b + y), 
and thus prove geometrically that the ^\P 

commutative and associative laws of ad- ^ 

dition hold true for complex numbers. 

MBeeCPee iy, OM ee a, 

MA eeDPee ib, MN = x. 

.-. (a + ib) + (x + m/) = J. + J.P = OP ; 
(a + iy) + (x + ib) = OS + BP = OP ; 
(a + x) + i (6 + y) = ON+ NP ee OP. 




EXERCISE XXX 

1. Prove (cos + ism 0) 3 =+ 1, 

[cos (2 Tzr/3) + i sin (2 tt/3)] 3 = + 1, 
[cos (4 tt/3) + i sin (4 tt/3)] 3 = + 1. 

(cos + i sin 0) 3 = cos (3 • 0) + i sin (3 • 0) = + 1. 
[cos (2 tt/3) + i sin (2 tt/3)] 3 = cos 2 it +.i sin 2 it ee =f- 1. 
[cos (4 7T/3) + i sin (4 tt/3)] 3 = cos 4 # -f i sin 4 it ee + 1. 
Hence each of these three quality units is a cube root of + 1. 

2. Prove [cos (ic/b) ± i sin (7T/5)] 5 = - 1, 

[cos (3 7T/5) ± i sin (3 tt/5)] 5 = - 1, 
(cos it ± i sin ?r) 5 = — 1. 
[cos (tt/5) ± i sin (7T/5)] 5 = cos # ± * sin tc = — 1. 
[cos (3 7r/5) ± sin (3 tt/5)] 5 = cos 3 % ± i sin 3 it 

ee cos tf ± i sin # = — 1. 
(cos Tt ± i sin 7r) 5 = cos 5 it ± i sin 5 it 

ee cos 7T ± i sin TT = — 1. 
Hence each base is a fifth root of — 1. 



EXERCISE XXX 



113 



3. Prove [cos (rt/6) ± isin(7T/6)] 6 

[cos (7T/6) ± i sin (tt/6)] 6 

[cos (tt/2) ± i sin (tt/2)] 6 

[cos (5 ?r/6) ± i sin (5 tt/6)] 6 



[cos (tt/2) ± i sin (tt/2)] 6 

[cos (5 tt/6) ± i sin (5 tt/6)]6 = - 1 . 

cos it ± i sin it = — 1. 

cos 37r±£sin37T = — 1. 

cos 5 7T ± £ sin 5 it = — 1. 

Illustrate the meaning of 




Fig. 11 



4. What is proved by examples 2 and 3 ? 
examples 1, 2, 3 by vectors. 

In example 2 it is proved that each base is a fifth root of — 1. 
In example 3 it is proved that each base is a sixth root of — 1. 

In fig. 11, 

OPi = cos + i sin = + 1. 
OP 2 = cos (2 7T/3) + i sin (2 tt/3). 
OP 3 = cos (4 7T/3) + i sin (4 jt/S) 

= cos (2 tz/3) - i sin (2 tt/3). 

If a unit line starting from OX is revolved 
through three times the angle XOPi, 
XOP 2 , or XOP 3 , its final position will be 
OX which represents + 1. This illustrates 
that the cube of any one of the quality units 
represented by these lines is + 1. 

Observe that OP 2 and OP 3 represent reciprocal or conjugate quality 
units. 

In fig. 12, OP 1 and OP 5 represent cos (n/b) ± i sin (iz/o) ; 

OP 2 and OP4 represent cos (3 nr/5) ± i sin (3 it/b) ; 
OP 3 represents cos it ± i sin #. 

If a unit line starting from OX is revolved 
through five times the angle XOPx, XOP 5 , 
XOP 2 , XOP±, or XOP s , its final position 
will be OP 3 , which represents — 1. This 
illustrates that the fifth power of any one 
of the quality units represented by these 
lines is — 1. 

Observe that OPi and OP 5 , or OP 2 and 
OP 4 , represent reciprocal or conjugate 
quality units. 

In fig. 13, OPi and OP & represent cos (tt/6) ± i sin (7t/6) ; 
OP 2 and OP 5 represent cos (tt/2) ± £ sin (tt/2) ; 
OP 3 and OP 4 represent cos (5 nr/6) ± i sin (5 ?r/6). 




114 



PLANE TRIGONOMETRY 



If a unit line starting from OX is revolved 
through six times the angle XOPi, XOPq, 
XOP 2 , XOP 5 , XOP 3 , or XOP±, its final 
position will be OX\ which represents — 1. 
This illustrates that the sixth power of any ^ 
one of the quality units represented by these X 
lines is — 1. 

Observe that OP x and OP 6 , OP 2 and OP 5 , 
or OP 3 and OP4, represent reciprocal or 
conjugate quality units. 



_ _ (cos 6 + i sin 0) 6 . 

5. Express — as a quality unit in the type form. 

(cos — i sin 0) 5 

(cos + i sin 0) 6 _ (cos + i sin 0) 6 




(cos <p — i sin 0) 5 [cos ( — 0) + i sin ( — 0)] 5 
(cos 6 -f i sin 6 0) 
cos ( — 5 0) + z sin ( — 5 0) 
= (cos 6 + i sin 6 0) (cos 5 + i sin 5 0) 
= cos (6 + 5 0) + i sin (6 + 5 0). 



6. Express 
(cos 



(cos0 — isin0) 
(cos0 — i sin 0) s 
isin6»)2 _ [cos(— 0) 



- in the type form. 
i.sm(- 0)] 1 



(cos0 — i sin 0)§ [cos (— 0) + isin (— 0)]* 
_ cos (- 3 0/2) + i sin (- 3 0/2) 
~ cos ( - 5 0/3) + i sin ( - 5 0/3) 

= [cos ( - 3 6/2) + i sin ( - 3 0/2)] [cos (5 0/3) + i sin (5 0/3)] 
= cos (5 0/3 -3 0/2) + i sin (5 0/3 - 3 0/2). 



7. Express 



(cos + i sin 0) (cos + i sin 0) 



in the type form. 



(cos p + i sin /3) (cos 7 + i sin 7) 

(cos + i sin 0) (cos + i sin 0) _ cos (0 + 0) + i sin (0 + 0) 
(cos p -f i sin /3) (cos 7 + i sin 7) cos (/3 + 7) + i sin (/3 + 7) 
= [cos(0 + 0) + isin(0 + 0)][cos(— |8 - 7) + isin(— p 
= cos (0 + - p - 7) + i sin (0 + - p - 7). 



7)] 



EXERCISE XXXI 115 



„ „ [cos (7T/6) - i sin (tt/6)] 2 " . x1 

8. Express t *-^ i m the type form. 

[cos (7T/6) + i sin (7t/6)]* 

[cos (tt/6) - t sin (rc/6)] *~ _ [cos ( - tt/6) + i sin ( - ^r/6)]' 
[cos (it/6) + i sin (7r/6)]* ~ [cos (tt/6) + i sin (tt/6)]* 

_ [cos(-ll7r/12)+isin(-ll^/12)] 
cos (7T/12) + i sin (7T/12) 

= cos (— it) + % sin (— nr) = — 1. 



. (cos + i sin 0) 4 ."-.,. - 

9. Express in the type form. 

(sin 6 + i sin 0) 5 

(cos + i sin 0) 4 _ cos40+zsin4 



(sin + i cos 0) 5 [cos (ar/2 — d) + i sin (?r/2 - 0)] 5 
_ cos 4 + i sin 4 

— cos (5 7i/2 - 5 0) + i sin (5 7T/2 -5 0) 
= cos (4 + 5 - 5 n/2) + i sin (4 + 5 - 5 m/2) 
= cos (4 + 5 - 7T/2) + i sin (4 + 5 - n/2). 



EXERCISE XXXI 

1. Find all the values of l 1 ' 3 . 

I 1 /3=(cos0 + isin0) 1/3 

= cos(2u7r/3) + isin(2n7r/3), where n = 0, 1, 2. 
When n = 0, = cos + i sin = 1. 

" n=l, =cos(2tt/3) + zsin(2 7T/3) = (-1 + i V3)/2. 

« n = 2, = cos (4 7T/3) + i sin (4 tt/3) = (- 1 - i V3)/2. 

2. Find all the values of (- l) 1 ' 3 . 

(_ 1)1/3 = ( C0S % _|_ i s i n ^)l /3 

2n7t-\-7t . . 2mt+7t , „ ^ 

= cos h*sm 5 where n = 0, 1, 2. 

3 3 

When n = 0, = cos (tt/3) + i sin (tt/3) = (1 + V^H)/2. 

" 7i = l, = cos 7r + i sin it = — 1. 

" n = 2, = cos (5 ?r/3) + i sin (5 ar/3) = (1 - V^3)/2. 



116 PLANE TRIGONOMETRY 

3. Eind all the values of l 1 ' 6 . 

li/6 =(cosO + i sin 0) 1/6 

7l7t 7l7t 

= cos — + i sin — , where n = 0, 1, 2, 3, 4, 5. 
3 3 

When n = 0, = cos + i sin = + 1. 

n = 1, = cos (n/S) + i sin (tt/3) = (1 + i V3)/2. 

n = 2, = cos (2 tt/3) + i sin (2 tt/3) = ( - 1 + i V«3)/2. 

n = 3, = cos # + i sin # = — 1. 

n = 4, = cos (4 tt/3) + i sin (4 tt/3) = (- 1 - « V3)/2. 

n = 5, = cos (5 nr/3) + i sin (5 tt/3) = (1 - i V3)/2. 

4. Eind all the values of ( — l) 1 /6 . 

(— l)i/6— (cos7T + 'isin7r) 1/6 

2mt-\-it . . 2n7t-\-7t 

= cos \- 1 sin , 

6 6 

where n = 0, 1, 2, 3, 4, 5. 

When n = 0, = cos (rt/6) + i sin (#/6) = ( V3 + i)/2. 

" 7i=l, = cos (7r/2) + i sin (tt/2) = + i- 

» n = 2, = cos (5 tt/6) + S sin (5 ?r/6) = (- V3 + i)/2. 

" ?i = 3, = cos (7 tt/6) + i sin (7 tt/6) = ( - V3 - i)/2. 

" n = 4, = cos (3 7T/2) + i sin (3 it/2) = — i. 

" n = 5, = cos (11 tt/6) + i sin (11 ?r/6) = (V3 - i)/2. 

5. Find all the values of 16i/ 4 . 

(+16)i/ 4 = (+1) 1/4 - (16)1/4 = (cosO + isin 0)i/4 x 2 . 
(cos 4- i sin 0) 1 /4 = cos \- i sin — , where n = 0, 1, 2, 3. 

When n = 0, = cos + i sin = -f 1. 

" n = 1, = cos (7t/2) + i sin (tf/2) = + i. 

" 71 = 2, = cos # + isin 7r = - 1. 

" 7i = 3, = cos (3 7t/2) + i sin (3 it/2) = - i. 

Hence the roots are ± 2, ± i • 2. 

6. Find all the values of 32 1 / 5 . 

(+32)1/5 = (+1)1/5. 32i/ 5 = (cos + isin0)i /5 .2. 
(cos + i sin 0)i /5 = cos (2 nit/b) + i sin (2 nit/b), where n = 0, 1, 2, 3, 4. 
When 7i = 0, = cos + i sin = 1. 

" n = l, =cos(2^/5)+isin(2^r/5). 

" 7i = 4 , = cos (8 x/b) + i sin (8 it/b) = cos (2 ?r/5) - i sin (2 it/b) . 

" 7i = 2, =cos(4 7r/5)+isin(4 7r/5). 

" n = 3, = cos (6 7T/5) + i sin (6 x/5) = cos (4 n/b) - i sin (4 n/b). 



EXERCISE XXXI 117 

Observe that the last four roots are reciprocals in pairs. 
Hence the five fifth roots of + 32 are 

2, [cos (2 rnc/b) ± i sin (2 titt/5)] • 2, where n = 1 or 2. 

7. Eind all the values of (- 243) 1 ' 5 . 

(-243)!/ 5 = (- l) 1 /5-243 1 ^ = (cos tt + isin tt) 1 ' 5 • 3. 
(cos it + i sin 7zr) 1/5 = cos [(2 n + 1) tt/5] + i sin [(2 n + 1) tt/5], 
where n = 0, 1, 2, 3, 4. 
The root is — 1 when n = 2. 
Hence the five fifth roots of — 243 are 



- 3, [cos (2 n + 1 tt/5) + i sin (2 n + 1 tt/5)] • 3, where n = 0, 1, 3, 4. 

8. Find all the values of (- i) 1 ' 6 . 
(_i)i/6 = [cos(- tt/2) + isin(- tt/2)]i /6 

= cos (4 n — 1 tt/12) + i sin (4 ?i — l tt/2), where n = 0, 1, 2, 3, 4, 5. 
Here the roots are opposites when n = and 3, 1 and 4, or 2 and 5. 
Hence 
(_ i)i/6 _ _j_ [ cos (4n-l tt/12) + i sin (4n-l tt/12)], where n = 0, 1, or 2. 

9. Eind all the values of (4 V3 + t 4) 1 ' 3 . 

(4 V3 + *4)i'* = (V3/2.+-'i/2) 1/8 -'2 = [cos (tt/6) + i sin (tz/G)] 1 '* • 2. 
[cos (?r/6) + i sin (tt/6)] 1 ' 3 

= cos (12 ft+1 tt/18) +£ sin (12n+l tt/18), where ?i= 0, 1, 2. 
Hence the three third roots are 



[cos (12 n + 1 tt/18) + i sin (12 n + 1 tt/18)] • 2, where n = 0, 1, or 2. 

10. Find all the values of (1 - i V3) 1 ' 4 . 

(1 - % V3p4 = [1/2 + .«(- V3/2)]!/4 ^/ 2 

= [cos ( - tt/3) + i sin ( - tt/3)] 1 ' 4 ^/2. 

[cos ( - tt/3) + z sin ( - tt/3 ) ] i '* 

= cos(6n-l7r/12)+isin(6?i-l7r/12), 
where n — 0, 1, 2, 3. 
When n = 0, = cos ( - tt/12) + i sin (- tt/12). 

" n = 2, = cos (11 tt/12) + i sin (11 tt/12) 

= -[cos(- tt/12) + isin(- tt/12)]. 
•' n=l, = cos (5 tt/12) + i sin (5 tt/12). 

" n = 3, = cos (17 tt/12) + i sin (17 tt/12) 

= - [cos (5 tt/12) + t sin (5 tt/12)]. 
Since these roots are opposite in pairs, the four fourth roots are 
± [cos (6 n -1 tt/12) + i sin (6 n - 1 tt/12)] ^/2, where n = or 1. 



118 PLANE TRIGONOMETRY 

11. Find all the values of (1 + V^"T)i /«, 
(1 + V^T) 1/6 = (l/v / 2+i/V2) 1/6] v/2 = [cos(7r/4) + isin(7r/4)] 1 /6 12/2. 
[cos (tt/4) + i sin (tt/4)] 1 ' 6 . 

= cos (8n + 1 tt/24) + i sin (8 n + 1 tt/24), 
where n = 0, 1, 2, 3, 4, 5. 
When n = 0, = cos (tt/24) + i sin (tt/24) . 

;t » = 1, = cos (9 tt/24) + i sin (9 tt/24). 

■' n = 2, = cos (17 tt/24) + i sin (17 tt/24). 

;t n = 3, = cos (25 tt/24) + i sin (25 tt/24). 

•' ft = 4, = cos (33 tt/24) + i sin (33 tt/24). 

ft = 5, ■ = cos (41 tt/24) + i sin (41 tt/24). 

Here the roots are opposites when n = and 3, 1 and 4, or 2 and 5. 
Hence the six sixth roots of 1 + V— 1 are 



± [cos (8 ft + 1 tt/24) + i sin (8 n + 1 tt/24)] ^2, where n = 0, 1, 2. 

12. Find all the values of ( V-3 - vTi)2/5, 

(a /3- V3I)2/5 = [ V3/2 + i(- l/2)]2/5 i/4 

= [cos ( - 7T/0) + i sin ( - 7t/Q)f^ -^4. 
[cos(— tt/6) -f isin( — 7T/6)] 2/5 

= [cos (- tt/3 ) + i sin ( - TT /3)] 1 / 5 
= cos (6 ft — 1 tt/15) -f i sin (0 ?i — 1 tt/15), 
where ft = 0, 1, 2, 3, 4. 
When ft = 0, = cos(— tt/15) + isin( — tt/15). 

" ft == 1, = cos (5 tt/15) + i sin (5 tt/15). 

" ft = 2, = cos (11 tt/15) + fsin (11 tt/15). 

" ft = 3, = cos (17 tt/15) + i sin (17 tt/15). 

" ft = 4, = cos (23 tt/15) + i sin (23 tt/15). 

Hence the five fifth roots required can be written 

[cos(rTT/15) + £sin(r7r/15)]^/4, where r = - 1, 5, 11, 17, 23. 

0. (1) 



13. 


Solve the equation x i 


- X s + X 2 - 


x + 


Mu 


Itiply by x + 1, 


X 5 + 1 = 


0. 




.-. x = (-l) 1 / 5 = (cos 


tc 4- i sin tt] 


1/5 . 



.-. x = cos (2 ft + 1 tt/5) + i sin (2 n + 1 tt/5), where ft = 0, 1, 3, 4. 
.-. x = cos (tt/5) ± i sin (tt/5), cos (3 tt/5) ± ism (3 tt/5). 
The solution — 1 was introduced by multiplying (1) by x + 1. 

14. Solve the equation x 12 - 1 = 0. (1) 

Factor, (x 3 - 1) (x 3 + 1) (x 3 - *) (x 3 + i) = 0. 

Hence the twelve solutions of equation (1) are I 1 / 3 , ( — l) 1 /3 , i 1 /3 , ( — i) 1/s . 



EXERCISE XXXI 119 

By example 1, 1 1/3 = + 1, (- 1 ± i V3)/2. 

By example 2, (- l) 1 ' 3 = - 1, (1 ± i s/S)/2. 

ii/a = [cos(?r/2) + isin(7r/2)]i/ 3 

= cos (4 n + 1 7T/6) + i sin (4 n + 1 ?r/6), where n = 0, 1. 2. 
.-. ii/3— cos (?r/6) + isin(7r/6), cos(5 7r/6) + isin(57r/6), - i. 
(- i)i/3 _ [ C os(- tt/2) + isin(- tt/2)] 1 / 3 

= cos (4 n — 1 7T/6) + i sin (4 n — 1 nr/6), where n = 0, 1, 2. 
.-. (- i)i/3 _ cos(- 7T/6) +isin(- tt/6), £, cos(7tt/6)+ i sin (7 tt/6) 
= cos (7r/6) — i sin (7r/G), i, cos (5 ^/6) — i sin (5 n/Q). 
The last six roots are ± i and cos (mt/6) ± i sin (wr/G), where n — 1, 5. 

15. Solve the equation x 1 + £ 4 + £ 3 + 1 = 0. (1) 
Factor, (x 4 + 1) (x 3 + 1) = 0. 

Hence the seven solutions of equation (1) are (— 1) 1/4 and (— l) 1 ' 3 . 
(- l)i/4 _ (costt + isin n) 1 '* 

= cos (2)1 + 1 7t/4) + i sin (2n+l 7r/4), 
where ra = 0, 1, 2, 3. 

= cos(7r/4) -f i sin (tt/4). 

= cos (5 7T/4) + i sin (5 7T/4) 

= — [cos (7r/4) + « sin (tt/4)]. 

= cos (3 tt/4) + i sin (3 tt/4). 

= cos (7 7T/4) + i sin (7 7T/4) 

= - [cos (3 tt/4) + i sin (3 tt/4)]. 
Byexample2,(-l)!/ 3 = -l, {l±iy/Z)/2. 

16. Solve the equation x" + 1 = 0. (1) 
The seven solutions of equation (1) are the seven values of (— 1) 1/7 . 

(-l)i/-= (costt 4- isimt) 1 ^ 

= cos (2 n + 1 tt/7) + £ sin (2 n + 1 7T/7), 

where n = 0, 1, 2, 3, 4, 5, 6. 
= cos (tt/7) + i sin (tt/7). 
= cos (13 it 1*1) + i sin (13 it /I) 
= cos (7T/7) — i sin (tt/7). 
= cos (3 n/1) + i sin (3 tt/7). 
= cos (11 7r/7) + i sin (11 7T/7) 
= cos (3 tt/7) - i sin (3 7r/7). 
2,4, = cos(5 7T/7) ± *sin(5 7r/7). 

= cos it + i sin 7T = — 1 . 
Observe that the roots are reciprocals of each other when n — and 6, 
1 and 5, or 2 and 4. 

Hence (— l) 1 ' 7 = — 1, cos(ri7r/7) ± i sin (tmt/7), where n = 1, 3, 5. 



When 


n 


= 0, 


1 1 


?i 


= 2, 


« 


n 


= li 


" 


n 


= 3, 



When 


n = 


1 <■ 


71 = 6 


a 


71 = 1, 


" 


n = 5 


u 


n = 2 


" 


n = 3 



120 PLANE TRIGONOMETRY 

EXERCISE XXXII 

1. Express all the angles which are co-terminal with 45°. 
45° + n • 360° denotes all the angles coterminal with 45°. 

2. Express all the angles which are coterminal with 132°. 
132° + n • 360° denotes all the angles coterminal with 132°. 

3. Express all the angles which are coterminal with — 35°. 

— 35° + n • 360° denotes all the angles coterminal with — 35°. 

4. Express all the angles which are coterminal with — 100°. 

- 100° + n • 360° denotes all the angles coterminal with — 100°. 

5. Express all the angles which are coterminal with tt/6. 
7i/6 + n • 2 7t denotes all the angles coterminal with it/6. 

6. Find all the other trigonometric ratios of A, when sin A = 4/7. 
Since sin A is + , A is in the first or second quadrant. 

Let OP = 7, then MP = 4. .-. OM = ± V72 - 42 = ± V33. 

.-. esc A = 7/4, cos A = ± V33/7, sec A = ± 7/V33, 

tan A = ± 4/V33, cot A= ± V33/4. 

7. Find all the other trigonometric ratios of A, when tan^l = 3/2. 
Since tan A is + , A is in the first or third quadrant. 

Let MP = ± 3, then OM = ± 2. .-. OP = V3 2 + 2 2 = y/lS. 

.-. cot A = 2/3, sin A = ± 3/V13, esc A = ± V13/3, 

cosJ. = ±2/V13, sec^l =±V13/2. 

8. Find all the other trigonometric ratios of A when cos A — - 3/8. 
Since cos A is — , A is in the second or third quadrant. 

Let OM = - 3, then OP = 8. .-. MP = ± V~8 2 - 9 = ± V55. 
.-. sec A = - 8/3, sin A = ± V55/8, esc A = ± 8/V&5, 
tan A = ± V55/( - 3) = T V55/3, cot ^ = =F 3/V55- 

9. Find all the other trigonometric ratios of A when cot A = - 7/5. 
Since cot A is — , A is in the second or fourth quadrant. 

Let OM = - 7, then MP = 5. .-. OP = V49 + 25 = V?4. 
Let OJf = 7, then MP = - 5. 

.-. tan ^4 = - 5/7, sin J. = ± 5/V74, esc A = ± V74/5, 
cos A = T 7/V74, sec A = T V74/7. 



EXERCISE XXXII 121 

10. Find all the other trigonometric ratios of A when sec A — 7/4. 
Since sec A is -f , A is in the first or fourth quadrant. 

Let OP = 7, then OM = 4, and MP = ± V49 - 16 = ± V33. 
.-. cos^l = 4/7, sin A = ± V33/7, esc A = ± 7/V33, 

tan A = ± V33/4, cot ^L = ± 4/V33. 

11. Find all the other trigonometric ratios of A when esc A = — 5/4. 
Since esc A is — , A is in the third or fourth quadrant. 

Let OP = 5, then MP = - 4, and Oil/ = =p Vo 2 - 4 2 = =f 3. 

.-. sin ^4 = — 4/5, cosJ.=t3/5, sec J. = T 5/3, 
tanJ. =-4/(^3) = ±4/3, cot^l=±3/4. 

12. In what quadrant is A in each of the examples 6-11 inclusive? 
Construct A in each case. 

The first part is answered under each example. 
The constructions are like those in Exercise VI. 

Find the value of the functions in examples 13-21 in terms of the 
functions of some positive angle less than 45°. 

13. sin 94° = cos 4°. 

14. cos 128° = - sin 38°. 

15. tan 215° = - cot 125° = tan 35°. 

16. cot 320° = cot (3 • 90° + 50°) = - tan 50° = - cot 40°. 

17. sec 190° = sec (2 • 90° + 10°) = - sec 10°. 

18. sin (- 75°) = - sin 75° = - cos 15°. 

19. cos (- 175°) = cos 175° = - sin 85° = - cos 5°. 

20. tan (- 200°) = - tan (2 • 90° + 20°) = - tan 20°. 

21. cot (- 300°) = - cot (3 • 90° + 30°) = tan 30°. 

22. Find the value of sin A in terms of each of the other functions of A. 

sin A = ± Vl - cos 2 ^., sin A = l/csc^4 = ± 1/Vl + cot 2 A, 

tan A tan A "Vsec 2 A — 1 

sin A = = ± = ± ' 

sec A Vl + tan 2 J. sec ^- 

23. Find the value of cos A in terms of each of the other functions of A. 

cos A = ± Vl — sin 2 J., cos A = 1/secA =± 1/Vl + tan' 2 A, 

cot A cot A Vcsc 2 ^. - 1 

cos A = - = ±— = =± 

cscA Vl + cot^. csc ^- 



122 PLANE TRIGONOMETRY 

24. Find the value of tan A in terms of each of the other functions of A . 



tan A = ± Vsec 2 -4 - 1, tan A = 1/cot A=± 1/Vcsc 2 A — 1, 

sin A sin A Vl — cos 2 ^i 

tan A = = ± = =± — 

cos A vl — sin 2 ^. cos .4 

25. Find the value of cot A in terms of each of the other functions of A. 



cot^l = ± Vcsc 2 ^. -1, cot J. = 1/tan ^4 = ± 1/Vsec 2 ^. - 1, 

cos A cos A Vl — sin 2 ^4 

cot^4 = - — - = ±—— — =±. . — .. 

smi Vl-cos 2 J. smA 

26. Find the value of sec A in terms of each of the other functions of A. 

sec A = 1/cobA-=± 1/Vl - sin 2 A, sec A = ± Vl + tan 2 A, 

1 esc A Vl+cot 2 J. _ , csc ^ 

sec A = = = ± 



cos A cot A cot A Vcsc 2 ^. — 1 

27. Find the value of esc A in terms of each of the other functions of A. 



esc A = 1/sin A = ± 1/Vl — cos 2 ^4, esc A = ± Vl + cot 2 A, 

1 sec^l Vl + tan 2 .4 secvl 

esc A = = = ± ■ = ± 



sin A tan A tan A Vsec 2 ^. — 1 

Identities 

28. Prove (tan A + cot A.) sin A cos A = l. 

(tan A + cot A) sin A cos A = sin 2 A + cos 2 A = l. 

29. Prove (sec A — tan A) (sec A + tan A ) = 1. 

(sec A — tan^.) (sec A + tan A) = sec 2 J. — tan 2 J. = 1. 

30. Prove (esc A — cot A) (esc A-+ cot A) = 1. 

(esc J. — cot J.) (esc J. + cot -4)= csc 2 ^L — cot 2 A = 1. 

31. Prove (sin B - cos I?) 2 = 1 - 2 sin E cos 5. 

(sin 5 - cos B) 2 = sin 2 B + cos 2 5 - 2 sin B cos J5 
= 1 — 2 sin 5 cos B. 

32. Prove sin B + cosB=^/2 cos (5 - tt/4). 

V2 cos (B - 7t/4) = V2 [cos 5 cos (tt/4) + sin 5 (sin ar/4)] 
= cos B + sin 5. 

33. Prove sin B - cos B = — V2 cos (2? + 7r/4). 

- V2 cos (5 + tt/4) = - V2[cos 5 cos (tt/4) - sin 5 sin (tt/4)] 
= sin 5 — cos jB. 



EXERCISE XXXII 123 

34. Prove sin (A + 7r/3) + sin (A — 7T/3) = sin 4. 
sin (A + tt/3) + sin (A - x/3) 

= sin J. /2 + cos A sin (tt/3) + sin A/2 — cos ^4 sin (7r/3) = sin J.. 

35. Prove cos (^4 + 7t/6) + cos (^4 — tt/6) = V3 cos -4. 
cos (J. + ar/6) + cos (A - ?r/6) 

= -y/3 cos A/2 —sin A sin (7t /6) + V3cos-4/2+sin^lsin (n/Q) = ye&cos A. 

36. (cot A + tan 5)/(tan A + cot 5) = cot A tan 5. 

(cot A + tan 5)/(tan A + cot 5) = cos A sin 5/(sin .4 cos 5) 

= cot J. tan 5. 

37. Prove 1 — tan 4 A = 2 sec 2 A — sec 4 A. 

(1 - tan 4 A) = (l-f tan 2 ^4) (1 - tan 2 ^.) = sec 2 ^4[l - (sec 2 J. - 1)] 

= 2 sec 2 A — sec 4 ud. 

38. sec 5/(1 + cos B) = (tan B - sin 5) /sin 3 B. 

(tan 5 — sin 5)/sin 3 B = (sin B — sin 5 cos B) /(cos 5 sin 3 5) 
= (1 - cos 5)/[cos 5 (1 - cos 2 5)] 
= 1/[cos5(1 + cos5)] 
= sec B/(l + cos B). 

39. Prove sec 2 ^4 esc 2 A = tan 2 ^4 + cot 2 A + 2. 

o . sin 2 ^4 cos 2 A 

tan 2 J. + cot 2 A + 2 — h \- 2 

cos 2 ^. sin 2 ^l 

= (sin 4 A + cos 4 A + 2 sin 2 J. cos 2 ^4)/(sin 2 ^4 cos 2 A) 

= (sin 2 A + cos 2 A) 2 /(sin 2 A cos 2 J.) 

= sec 2 A esc 2 A. 

40. Prove tan B + sec 5 = tan (5/2 + tt/4). 

tan (5/2 + tt/4) = [1 + tan(5/2)]/[l - tan (5/2)] 

= [cos (5/2) + sin(7V2)]/[cos(5/2)'- sin (5/2)] 
= [cos (5/2) + sin(5/2)] 2 /[cos 2 (5/2) - sin 2 (5/2)] 
= [1 + 2 cos (5/2) sin (5/2)]/cos 5 
= (1 + sin 5)/cos 5 
= sec 5 + tan 5. 

41. Prove (1 + tan 5)/(l - tan 5) = (cot 5 + l)/(cot 5 - 1). 
(1 + tan 5)/(l - tan 5) == (cos 5 + sin 5)/(cos 5 - sin 5) 

ee (cot 5 + l)/(cot 5-1). 

42. Prove sin A/(l + cos ^4) + (1 + cos A)/sin A = 2 esc A. 

sin A/(l + cos J.) + (1 + cos A)/sin A 

= (sin 2 J. + 1+2 cos ^4 + cos 2 ^4)/[sin^4 (1+ cos ^4)] 
= 2/sin A = 2 esc A. 



124 PLANE TRIGONOMETRY 

43. Prove sec- 3 .A - sin 3 A = (cos A - sin J.)(l + sin A cos A). 
sec- 3 ^. — sin 3 A = (cos 3 Jl - sin 3 .4) 

= (cos A - sin A) (cos 2 A + cos A sin A + sin 2 .A) 
= (cos A — sin A) (1 + cos A sin A). 

44. Prove (sin A cos B -f cos A sin B) 2 + (cos A cos 5 - sin A sin B) 2 = 1. 
(sin A cos B + cos ^4 sin J3) 2 + (cos J. cos B - sin J. sin B) 2 

= sin 2 ^4 cos 2 B + 2 sin .A cos 5 cos A sin .B + cos 2 ^4 sin 2 5 

+ cos 2 A cos 2 B - 2 cos A cos £ sin ^. sin B + sin 2 J. sin 2 £ 
EEcos 2 J3 + sin 2 £ = l. 

45. Prove cot A — tan A = 2 cot 2 J.. 

2 cot 2 J. = (cot 2 J. - l)/cot J. = cot A — tan J.. 

46. Prove sec 2^4 = sec 2 A/ (2 - sec 2 J.). 

sec 2 A = l/cos2 A = 1/(2 cos 2 J. - 1) = sec 2 A/(2 - sec 2 A). 

47. Prove 2 sec 2 A = sec (^4 + tt/4) sec (J. - tt/4). 
sec (J. + tt/4) sec (A - n/4) 

= 1/[| V2 (cos J. - sin ^4) | V2 (cos J. + sin A)] 

= 2/(cos 2 A-sh\ 2 A) 
= 2/cos 2 ^4 = 2 sec 2 A. 

48. Prove sin 2 A = 2 tan ^4/(1 + tan 2 ^4). 

2 tan J./(l + tan 2 J.) = 2 sin A cos ^4 = sin 2 J.. 

49. Prove 2 sin A + sin 2 J. = 2 sin 3 J./(l - cos A). 

2 sin 3 A/ (I - cos -4) = 2(1 - cos 2 A) sin ^4/(1 - cos A) 

= 2 sin A + 2 sin A cos J. = 2 sin ^4 + sin 2 A. 

50. Find sin (^4 + B + C) in terms of the sine and cosine of A, B, C. 
Applying formula [7] twice and [8] once, we obtain 

sin(J. + B + C) = sin (A + £)cosC + cos(J. + £)sin C 
= (sin A cos B + cos A sin 5) cos C 

+ (cos ^4 cos B — sin J. sin B) sin C 
= sin A cos I? cos C + cos J. sin B cos C 

+ cos A cos J5 sin C — sin ^4 sin B sin C. (1) 
If A, B, C are the angles of a triangle, sin (A + B + C) = sin 180° = 0. 
Hence, from (1), we obtain 
sin A cos B cos C + cos J. sin B cos C + cos ^4 cos 5 sin C = sin A sin 5 sin C. 



EXERCISE XXXII 125 

51. Applying formula [8] twice and [7] once, prove 

cos ( A + B + (7) = cos A cos B cos C — cos A sin B sin C 

— sin A cos B sin C — sin A sin B cos C. (2) 
cos (4 + B + C) = cos (A + 5) cos C - sin (A + B) sin C 

= (cosA. cosB — sin^i sin5) cos C — (sinA cosB + cosA sinB) sin C 
= cos A cos 5 cos C — sin J. sin B cos C — sin A cos 5 sin C 
— cos A sin B sin C. 

52. Find tan (A + 5 + C) in terms of tan J., tan 5, tan C. 
Applying formula [11] three times, we obtain 

tan (A + B) + tan O 



tan ( J. + 5 + C) 



1 - tan {A + £) tan C 
tan A + tan _B 



tan A tan J5 



tan C 



tan A. + tan B J _. 

1 tan C 

1 — tan ^i tan 5 



_ tan A + tan B + tan C — tan A tan i? tan C 
1 — tan J. tan B — tan A tan C — tan 5 tan C 
HA,B,C are the angles of a triangle, tan (A + B + C) = tan 180° = 0. 
Hence, from (3), 

tan A + tan B + tan C = tan A tan 5 tan C. 

53. Putting A = B = C in (1), (2), (3) of examples 50, 51, 52, prove 

sin 3 A = 3 sin A(l — sin 2 A) — sin 2 J. 

= 3 sin A. -4 sin 3 A, (4) 

cos 3 A = 4 cos 3 J. — 3 cos A, (5) 

_ 3 tan A - tan 3 A. „ 

tan 3 A = (6) 

l-3tan 2 A 

Putting A = B = C in (1) of example 50, we obtain 

sin (3 ^4) = 3 cos 2 ^4 sin A — sin 3 J. 

= 3 sin A (1 — sin 2 A) — sin 3 A 

= 3sin^l — 4 sin 3 A. 

Putting A = B = C in (2) of example 51, w T e obtain 

cos (3 A) = cos s A — 3 cos A sin 2 A 

= cos 3 vl — 3 cos A{\ — cos 2 A) 

= 4 cos 3 A — 3 cos A. 

Putting A = B = C in (3) of example 52, we obtain 

tan A — tan 3 A 



tan 3 A 



1 -3 tan 2 A 



126 PLANE TRIGONOMETRY 

54. Writing 3A = 2A-\-A, prove (4), (5), (6) in example 53 by using 
[7], [8], [11], [13], [14], and [15]. 
sin 3 A = sin (2 A + A) 

= sin 2 A cos A + cos 2 A sin A 

= 2 sin A cos ^4 • cos A + (cos 2 -J. — sin 2 J.) sin A 

= 3 sin A (1 — sin 2 J.) — sin 3 J. 

= 3 sin A — 4 sin 3 J.. 
cos 3 A = cos (2 A + ^4) 

= cos 2 J. cos ^4 — sin 2 J. sin A 

= (cos 2 J. — sin 2 A) cos A — 2 sin A cos ^4 sin A 

= cos 3 ^4 — 3 cos A (1 — cos 2 A) 

= 4 cos 3 ^4 — 3 cos A. 
tan 3 A = tan (2 ^4 +^4) 

tan 2 J. + tan ^4 

~~ 1 — tan 2 A • tan ^4 

2 tan J. . 

+ tan J. 



1 - tan 2 ^4 



2 tan ^4 

1 tan A 

1 - tan 2 A 

_ 2 tan J. + tan A — tan 3 A 

~ 1 - tan 2 A - 2 tan 2 ^4 

_ 3 tan J. — tan 3 ^4 

l-3tan 2 J. 

55. Substituting for 3 A in (4) and (5) of example 53, we obtain 

sin = 3 sin (6/3) - 4 sin 3 (0/3), 
cos = 4 cos 3 (0/3) - 3 cos (0/3). 

56. Prove sin 4 J. = 4 sin A cos A — 8 sin 3 -4 cos ^4 

= 8 cos 3 v4 sin A — 4 cos A sin ^4. 
sin 4 J. = sin (^4 + 3^4) 

= sin A cos 3 A + cos -4 sin 3 ^4 

= sin A (4 cos 3 J. — 3 cos A) + cos A (3 sin J. — 4 sin 3 ^4) 

= 4 sin J. cos 3 ^4 — 3 sin .A cos J. + 3 sin J. cos A 
— 4 sin 3 A cos ^4 

== 4 sin A cos 3 ^4 — 4 sin 3 A cos J. 

= 4 sinJ. cos A (1 — sin 2 A) — 4 sin 3 ^4 cos J., 
or 4 sin ^4 cos 3 A — 4 sin ^4 cos ^4 (1 — cos 2 ^4) 

= 4 sin A cos J. — 8 sin 3 A cos J., 
or 8 sin A cos 3 J. — 4 sin J. cos -4. 



EXERCISE XXXII 127 

57. Prove cos 4 A = 1 - 8 cos 2 A + 8 cos 4 ^. 

= l-8sin 2 J.+ 8 sin 4 A. 
cos 4 A = cos (3 A + J.) 

— cos 3 ^4 • cos A — sinSA-smA 

— (4 cos 3 A — 3 cos J.) cos A — (3 sin A — 4 sin 3 .4) sin .4 
= 4 cos 4 J. - 3 cos 2 A - 3 sin 2 A + 4 sin 4 ^. 

= 4(1 - sin 2 J.) 2 - 3 (cos 2 J. + sinM) + 4sin 4 ^, 
or 4 cos 4 J. - 3 (cos 2 ^L + sin 2 ^4) + 4 (1 - cos 2 ^!) 2 

= l-8sin 2 J.+ 8 sin 4 .4, 
or 1 - 8 cos 2 J. + 8 cos 4 ^4. 

58. Prove cos 780° = 1/2. 

cos 780° = cos (2 • 360° + 60°) = cos 60° = 1/2. 

59. Prove sin 1485° = V2/2. 

sin 1485° = sin (4 • 360° + 45°) = sin 45° = V2/2. 

60. Prove cos 2550°= V3/2. 

cos 2550° = cos (7 • 360° + 30°) = cos 30° = V3/2. 

61. Prove sin (- 3000°) = - V3/2. 

sin ( - 3000°) = - sin (3000°) = - sin (8 • 360° + 120°) 

= - sin 120° = - cos 30° = - V3/2. 

62. Prove tan (- 2190°) = - V3/3. 

tan ( - 2190°) = - tan 2190° = - tan (6 • 360° + 30°) 
= - tan 30° = - V3/3. 

Equations 

63. In what quadrant is A, if sin A cos A — — 2/3 ? 

Since sin A cos A is — , sin J. and cos A are opposite in quality ; hence 
A is in the second or fourth quadrant. 

64. In what quadrant is A, if sin A tan A = 4 ? 

Since sin A tan A is + , sin A and tan A are like in quality ; hence A is 
in the first or fourth quadrant. 

65. In what quadrant is J., if sec A tan A = — 3 ? 

Since sec A tan A is — , sec A and tan A are opposite in quality ; hence 
A is in the third or fourth quadrant. 

66. In what quadrant is A, if cot A + 3 sin A = ? 

Since cot A + 3 sin A = 0, cot A and sin A are opposite in quality ; 
hence A is in the second or third quadrant. 



128 PLANE TRIGONOMETRY 

67. Solve the equation sin 2 = 2 cos 0. (1) 
Substituting for sin 2 its identical expression 2 sin cos 0, from (1) by 

Algebra we obtain the equivalent equation 

2 sin cos = 2 cos (9, or cos (sin — 1) = 0. 
From cos = 0, - 2 wtt ± jr/2. § 70 

From sin = 1, = ntf + (- 1) w tt/2. § 69 

Hence nit ± n/2 includes all the values of in (1). 

68. Solve cos 2 = 2 sin 0. 

cos2 0=l - 2sin 2 0. 
.-. l-2sin 2 = 2sin0. 

.-. sin0 = -.5 ±.5 V3 = -.5 ±.5(1.7321). 
.-. sin0 = .3661, or - 1.3661. 
From sin = . 3661 = sin 21° 28', 

0= n- 180° + (-!)* 21° 28'. 
sin = — 1.3661 is impossible, that is, no value of will satisfy this 
condition. 

69. Solve cos = sin 2 0. 

sin 2 = 2 sin cos 0. 
.-. cos — 2 sin cos = 0, 
or cos 0(1 - 2sin0) = 0. 

From sin = 1/2 = sin (ar/6), 

= nit + (— 1)»tt/6. 
From cos = = cos (tt/2), 

= 2 ft7T ± 7T/2, Or ft7T + 7T/2. 

70. Solve sin = cos 2 0. 

cos20 = l -2sin 2 0. 
.-. sin0 = 1 - 2sin 2 0, 
or (sin + 1) (sin -1/2) = 0. 

From sin = — 1 = sin (— rt/2), 

6 = nit — (- YfTt/% 

From sin = 1/2 = sin (tt/6), 

= n7t + (-l) n 7t/6. 

71. Solve tan J. tan 2 J. = 2. 

tan 2 J. = 2 tan J./(l - tan 2 A). 
.-. tan 2 4 = 1 - tan 2 A. 
.-. tan 4 = ± V2/2 = tan ( ± 35° 16'). 
.-. 4 = ft • 180° ± 3.5° 16'. 



EXERCISE XXXII 129 

72. Solve cos A + cos 2 A = 0. 

cos 2A = 2 cos 2 A — 1. 
.-. cos A + 2 cos 2 J. -1 = 0, 
or (cos J. - 1/2) (cos A + 1) = 0. 

From cos A = 1/2 = cos(7r/3), 

A=2mt ± 7T/3. 
From cos J. = — 1 = cos 7t, 

A = 2 rnt ± it. 

73. Solve cot A tan 2 ^i = 3. 

cot J. tan 2^4 = 2/(1 - tan 2 A). 
.-.2 = 3 -3tan 2 ^l. 
.-. tan A = ± V3/3 = tan (± tt/6). 

.'. A = Mt ± 7Z/Q. 

74. Solve 4 cos 2 J. + 3 cos A = 1. 

cos 2 J. = 2 cos 2 .4 — 1. 
.-. 8cos 2 J. + 3cosJ.- 5 = 0, 
or (cos A + 1) (cos A - 5/8) = 0. 

From cos A = — 1 = cos 180°, 

A = n • 360° ± 180°. 
From cos -4 = 5/8 = cos 51° 19', 

A = n- 360° ± 51° 19'. 

75. Solve sin sec 2 = 1. 

cos 2 = 1 - 2 sin 2 0. 
.-. sin = cos 2 = 1 - 2 sin 2 0, 
or sin 2 + (1/2) sin - 1/2 = 0. 

.-. (sin + 1) (sin - 1/2) = 0. 
From sin = — 1 = sin ( — 7t/2), 

6 = mt -(-l)"7f/2. 

From sin = 1/2 = sin (tf/6), 

= mt + (- 1)»tt/6. 

76. Solve cot tan 2 = sec 2 0. 

cot tan 2 = 2/(1 - tan 2 0) = 2 cos 2 J./ (cos 2 A - sin 2 4) . 

sec 2 = 1/cos 2 = l/(cos 2 ^L - sin 2 J.). 
.-. 2 cos 2 A = 1, or cos J. = ± V2/2. 
From cos J. = V2/2 = cos (tt/4), 

^. = 2 717T ± 7T/4. 
From cos A = — V 2 / 2 = cos (5 7T/4), 

A = 2 ?17T ± 5 7T/4 

= (2 W ± 1) 7T ± 7T/4. 

All these values of A are included in nit ± tt/4. 



130 PLANE TRIGONOMETRY 



77. 


Solve 


sin 


2 = 3 sin 2 


- cos 2 0. 








. 










sin 2 = 


2 sin cos 0. 












.-. 2 sin0 cos0 = 


3 


sin 2 - cos 2 0. 












•. 2 cot0 = 


3 


- cot 2 .0. 












(cot0 + 3)(cot0- 1) = 


0. 








From 






COt0 = 


- 


3 = cot(- 


18° 


260, 










= 


n 


180° - 18° 


26 / . 




From 






COt0 = 


1 


= cot 45°, 














= 


n 


180° + 45° 






78. 


Solve 


sin 


9 + cos2 = 

.-. sin0 + 1 
6sin 2 - 


4sin 2 0. 

cos 2 = 
-2sin 2 = 
sin - 1 = 


i 

4 

0. 


-2sin 2 0. 
sin 2 0, 







.-. (2 sin 0-1) (3 sin + 1) = 0. 
From sin = 1/2= sin 30°, 

0= ?il80° + (-1)»30°. 
From sin = - 1/3 = sin ( - 19° 28'), 

d = n 180°- (-T)»19°28'. 
79. Solve sin 2 = cos 4 0. 

cos 4 = 1 - 2 sin 2 2 0. 
.-. sin 2 =1 -2sin 2 2 0. 
.-. (sin 2 + 1) (sin 2 - 1/2) = 0. 
From sin 2 = — 1 = sin(— tt/2), 

e = mt/'2 _(_l)"7r/4. 
From sin 2 = 1/2 = sin (7T/G), 

0= nn/2 + (- 1)"tt/12. 
±V3. 



80. Solve sec x + tan x 


Multiply by 


cos X, 




.-. 2sin 2 x 


or 


(2 sin x — 1 


From 





1 + sin x = ± V3 Vl — sin 2 x. 
sinx — 1 = 0, 
.)(sinx + 1) = 0. 

sinx = 1/2 = sin (tt/6), 
x = mt + (- 1)"tt/6. 
From sin x = — 1 = sin ( — 7r/2), 

X = ?17T — (— l)"7T/2. 

Since sec x + tan x is finite, the second value of x does not satisfy the 
equation. These values were introduced by squaring. 



EXERCISE XXXII 131 

81. Solve tan x + 2 V3 cosx = 0. 

Since tan x + 2 y/S cos x = 0, tan x and cos x are opposite in quality ; 
hence x is in the third or fourth quadrant. 

sin x + 2 V3 cos 2 a; = 0, 
or sin x + 2 V3 (1 - sin 2 x) = 0. 

.-. sinx = - V3/2, or 2/V3. 
From sin x — — %/3/2 = sin ( — 7T/3), 

= n7t-{-l) n rt/S. 
sin x = 2/>/3 is impossible. 

82. Solve 2 sin 2 x - 2 = - V2 cosx. 

sin 2 x = 1 — cos 2 x. 
.-. — 2 cos 2 x = — V2 cos x, 
or cos x (2 cos x — y/2) = 0. 

From cos x = V2/2 = cos (tt/4), 

x ='2n7T ± 7T/4. 
From cosx = = cos(7r/2), 

x = 2 )17T ± 7T/2. 

83. Solve cos 2 y + 2 sin 2 ?/ - (5/2) sin y = 0. 

cos 2 y = 1 — sin 2 ?/. 
.-. 1 + sin 2 ?/ - (5/2) sin ?/ = 0, 
or (sin y — 2) (sin ?/ — 1/2) = 0. 

sin y = 2 is impossible. 
From sin ?/ = 1/2 = sin (n/G), 

y = nit'+(— l) n 7t/6. 

84. Solve sin + sin 2 = 1 - cos 2 0. (1) 
From (1), ' sin + 2 sin cos = 2 sin 2 0. 

.-. sin = 0, or 1 + 2 cos = 2 sin 0. 
From sin = = sin 0°, 

= n ■ 180°. 

Also 1 + 2 cos = 2 sin = 2 Vl - cos 2 0. (2) 

.-. 1 + 4 cos + 4 cos 2 = 4-4 cos 2 , 
or cos 2 + 1/2 cos = 3/8. 

.-. cos = - 1/4 ± V7/4 = - .25 ± .6614 = .4114, or - .9114. 
From cos = . 41 14 = cos 65° 42' 20", 

= n 360° + 65° 42' 20". 
From cos = - .9114 = cos 204° 18", 

= n 360° + 204° 18'. 
From (2), sin — cos < 1 numerically, whence sin and cos are like 
in quality ; hence is in the first or third quadrant. 



132 PLANE TRIGONOMETRY 

85. Solve cosy — cos 2y = 1. 

cos2y = 2 cos 2 y — 1. 
.-. cos y — 2 cos 2 ?/ + 1 = 1, or cos y (2 cos y — 1) = 0. 
From cos y = = cos (7T/2), 

y = 2 717T ± ?r/2. 
From cosy = 1/2 = cos (7T/3), 

y = 2 n# ± 7T/3. 

86. Solve sin (45 + z) + cos (45 - z) = 1. 
By [7] and [10], we obtain 

sin 45 cos z + cos 45 sin z + cos 45 cos z -f sin 45 sin z = 1. 

.-. cos z + sin z = V2/2. (1) 

Since cos z + sin z lies between and +1, cos z and sin z are opposite in 
quality and the positive function is the greater arithmetically ; hence z 
is in the second or fourth quadrant. 
Squaring (1), 

cos 2 z + sin 2 z + 2 sin z cos z = 1/2. 
.-. 2 sin z cos z = — 1/2, 
or sin2z = - 1/2 = sin (-30°). 

.-. 2z = nl80° + (-l)»(-30°), 
.-. z = n90°- (-1)^15°. 

When n = 0, z = - 15° ; when n = 1, z = 105° ; when n = 2, z = 165°. 
But (1) is satisfied only by the angles coterminal with — 15° or 105°; 
hence z = n- 360° - 15°, or n ■ 360° + 105°. 

87. Solve sec 2 z + 1 = 2 cos z. 

1/cos 2 z + 1 = 2 cos z. 
.-. 1 + cos2z = 2 cosz • cos2z. 

1 + cos2z = 2cos 2 z, cos2z = 2cos 2 z — 1. 
.-. 2 cos 2 z = 4 cos 3 z — 2 cosz, 
or cosz(— 2 cosz + 4 cos 2 z — 2) = 0. 
.-. cos z (2 cos z + 1) (2 cos z — 2) = 0. 
From cos z = = cos (tt/2), 

z = 2n7t ± 7r/2. 
From cos z = — 1/2 = cos (2 7T/3), 

z = 2 nit ± 2 ?r/3. 
From cos z = 1 = cos 0°, 

z = 2 mt. 



EXERCISE XXXII 133 



88. Solve cos 2 z — a (1 — cos z). 

cos 2 z = 2 cos 2 z — 1. 
.-. 2 cos 2 z — 1 = a — a cos 2, 
or cos 2 z -f - cos z — (1 + a)/2 = 0. 

.-. cos z = ( - a ± Va 2 + 8 a + 8)/4. 

, -ii n , - a ± Va 2 + 8 a + 8 
.-. 2; = 2 W7r ± principal value of cos -1 

89. Solve tan 2 y tan y = 1. 

tan 2 ?/ = 2 tan ?//(l — tan 2 ?/). 
.-. 2 tan 2 ?/ = 1 — tan 2 y. 

.-. tan y = ± V3/3 = tan (± ?r/6). 

.*. y = 7l7t ± 7t/6. 

90. Solve sec = 2 tan + 1/4. 

sec d = Vtan 2 + 1. 
.-. tan 2 + 1 = 4 tan 2 + tan d + 1/1G, 
or tan 2 9 + (1/3) tan d = 5/16. 

.-. tan B = 5/12, or - 3/4 = .4167, or - .7500. 
From tan 6 = .4167 = tan 22° 37' 20", 

6 = n 180° + 22° 37' 20 // . 
From tan 6 = - . 7500 = tan 143° T 48", 

6 = n 180° + 143° T 48". 

91. Solve sin-ice + sin- 1 (x/2) = 120°. 

cos [sin-ice + sin- 1 (x/2)] = cos 120° = - 1/2. 
Let A — sin - : x, 

and B = sin- 1 (x/2). 

Then sin^i = x, 

sin B = x/2, 



cos A = VT 



x z , 



cos E = V4 - x 2 /2. 



.-. cos (A + B) = Vl-x 2 V4 - x 2 /2 - x 2 /2 = - 1/2. 

.-. (1- X 2 )(4 - X 2 ) = (X 2 - l) 2 . 

.-. x = ± 1. 



134 PLANE TRIGONOMETRY 



92. Solve 


sin~ 


1 z + 2cos- 1 « = 210 









Let 




A = 
73 = 


sin- 1 z, 

COS -1 2. 






Then 




sin (A + 2 B) = 
sin A = 
cos A = 
cosB = 
sin B = 
.-. cos 2 73 = 


sin 210° = -1/2, 

Vl - 2 2 , 

Z i 

Vl - jga. 
= 2 cos 2 73-1 = 2z 2 -l 


5 








sin 2 73 = 


:2sin73cos73=2z Vl 


-z 2 . 








sin (A + 2B) = 


= sin A cos 2 B + cos J. 


sin 2 73. 




z(2z 2 -l) 


+ 2 


Wl-2 2 Vl - Z 2 = 
.'. Z = 


-1/2. 
-1/2. 






93. Solve 


tan~ 


1 y + 2 cot- 1 y = 13£ 


°. 






Let 




^4 = 
73 = 


tan- 1 y, 
cot - 1 y. 






Then 




tan (A + 2B) = 

tan J. = 

.-. tan 7? = 


tan 135° = - 1, 
y, cot B = y. 

ir 1 , 










tan 2 B = 


2 tan 73/(1 - tan 2 73) = 


2y/(y*- 


1). 






tan (i + 25)E 


(tan A + tan 2 73)/(l - tan ^1 tan 2 


£) 






- (y B + y)/{y* + l) = 


-1. 










••• 2/ = 


1, ±i. 










K- = 


± i is not admissible. 










2z 









94. Solve tan- 1 — 
1- 


= 60°. 

-Z 2 




2 2/(1 - z 2 ) = tan 60° = V3 
.-. 2 2 = V3 - V3 z 2 . 
.-. z = V3/3, - V3. 



95. Solve tan- 1 z + tan- * 2 z = tan- i 3 VS. 
Let A = tan _1 z, 

73 = tan- 1 2 z. 
Then tan (A + 73) = 3 V3, 

tan ^4 = z, tan 73 = 2 z. 
tan (^L + 73) = (tan J. + tan 73)/(l - tan A tan 73). 

.-. (z + 2z)/(l-2z 2 ) = 3V3. 
.-. 2 V3z 2 + z = V3. 

.-. z = V3/3, - V3/2. 



EXEKCISE XXXII 135 



96. 


Solve tan x + tan 2 


c = 0. 














tan 2 x e 


E2tan 


x/(l- 


- tan 2 x). 




.-. tanx + - 


2 tan x 
— tan 2 x 


= 0. 










.-. tanx (3 


- tan 2 x) = 


= 0. 








From 




tanx = 


= = tan 0°, 










x = 


= mt. 








From 




tanx = 

x = 


= mt ± 


= tan 

It/%. 


(± 


7T/3) 



97. Solve tan 2 x + cot 2 x = 10/3. 

tan 2 x + l/tan 2 x = 10/3. 
.-. tan 4 x - (10/3)tan 2 x = - 1. 

.-. tan 2 x = 3, or 1/3. 
From tan x = ± y/S = tan ( ± tt/3), 

x = nit ± if/3. 
From tan x = ± V3/3 = tan ( ± 7t/6), 

x = nit ± it/6. 

98. Solve 4 cos 2 + 6 sin 6 = 5. 

4(1- 2sin 2 0) + 6sin0 = 5. 
.-. (sin 0-l/2)(sin 0-1/4) = 0. , 
From sin = 1/2 = sin (tt/6), 

6 = nit + (-l) n 7t/6. 
From sin = .25 = sin 14° 28' 40", 

= ?i 180° + (- 1)* 14° 28' 40". 

99. Solve sin A + cos A = sec A. 

Multiply by cos A, sin A cos A = 1 — cos 2 J. = sin 2 A. 

.-. sin J. (tan A — 1) = 0. 
From sin X = = sin 0, 

A — nit. 
From tan A = 1 = tan (7T/4), 

^4 = ?17T + 7T/4. 

100. Solve sin (0 + 30°) sin (0 - 30°) = 1/2. 

[sin ( V3/2) + cos 0/2] [sin ( V3/2) - cos 0/2] = 1/2. 

(3/4) sin 2 - cos 2 0/4 = 1/2. 
(3/4) sin 2 - (1 - sin 2 0)/4 = 1/2. 

.-. sin = ± V3/2 = sin (± 7t/S). 

.-. = nit + (- l) n (± 7T/3) = nit ± 7t/S. 



136 PLANE TRIGONOMETRY 





Systems 


OF 


Equations 


101. 


Solve for r and the system 








r sin = 


: a, 


(2)/ 






r cos = 


:'&. 




Divide (1) by (2), tan = 


:a/6, 


or = tan~ 


i (a/6). 


Square (1) and (2) and add, 










r 2 (sin 2 + cos 2 0) = 


: a 2 + 6 2 . 






.-. r = 


;Va 2 


+ 6 2 . 




102. 


Solve for r, 0, and the system 






r cos sin = 


:a, 


(1) ] 






r cos cos - 


:■&, 


(2)t 





(a) 



(a) 

r sin = c. (3) J 
Divide (1) by (2), tan = a/b, or = tan- i (a/6). (4) 

Square (1) and (2) and add, 

r 2 cos 2 = a 2 + 6 2 . (5) 

From (5), r cos = Va 2 + 6 2 . (6) 

Divide (3) by (6), tan = c/Va 2 + 6 2 , or = tan- 1 (c/VaM^6~ 2 ). 

Square (3) and add (5), r = Va 2 + 6 2 + c 2 . (7) 

103. Solve for x and ?/ the system 

sin x + sin y = a, (1)\ 
.cosx + cosy = 6. (2)1 
By § 40 we obtain from system (a) the equivalent system (b). 
2sml(x + y)cosl(x-y) = a, (3)| 
2 cos \(x + y) cos i (a; — y) == 6. (4) J 
Divide (3) by (4), tan | (x + y) = a/6. (5) 

Hence sin j (x + ?/) = ± a/Va 2 + 6 2 . (6) 

Substituting the value of sin i(x + y) in (3), we obtain 

cos \(x-y) = ± Va 2 + 6 2 /2. (7) 

From (5), (x + y)/2 = tan- 1 (a/6). (8) j 

From (7), (x - y)/2 = cos-i (± Va 2 + 6 2 /2). (9) J 



Hence x = tan- 1 (a/6) + cos- 1 ( ± Va 2 + 6 2 /2), 

and y = tan- 1 (a/6) - cos-i(± Va 2 + 6 2 /2). 



EXERCISE XXXII 137 

104. Solve the system 



sin x — sin y = a, 1 

J 



cosx — cosy = b. „ 
cos i (x + y) sin | (x - y) = a/2, (in 
sin i(x + y) sin ±(x -y)=- b/2. (2) J* 

.-. tan|(x + y) =- b/a, or I (x + ?/) = tan-!(- 6/a). i 

.-. sin i (x + y) = ± b/Va? + 6 2 . }> 

.-. sin l(x-y) = Wa 2 + & 2 /2, or i(x-y) = s in-i(TV a 2 +&72). J 
.-. x = tan-i(-6/a) + sin-i(T^ / a^T^/2),| 
y = tan-i(- 6/a) - sin-^T Va 2 + 6 2 /2). J 

105. Solve for x and y the system 

x sin A + ?/ sin U = a, (1) "1 
xcos^l + ycosB = b. (2) J 
From (2), x = (b - y cos 5) /cos A. (3) 

From (1), (3), 

(6 sin ^1 — ?/ sin A cos 5) /cos J. + y sin 5 = a. 

.-. y — {a cos vl — 6 sin .A)/(sin 5 cos A — cos B sin A) 
= (a cos A — b sin ^l)/sin (5 — A) . 
Similarly, x = (6 sin 5 — a cos B) /sin (B — 4). 



106. Solve for r and 6 the system 

r sin (d + A) = a,\ 
rcos(d + JB) = 6. J 

By T71 and [8], from (a) we obtain the equivalent system (b). 
r sin cos A + r cos sin J. = a, "1 



(a) 



(b) 



J (o) 



r cos cos 5 — r sin sin 5 = b. J 
Solving (b) as a linear system in r sin and r cos as the unknowns, we 
obtain r sind = (acosB - bsmA)/cos(A - B), (1)\ 
r cos 6 = (b cos A + a sin B) /cos (J. — 5). (2) 
.*. tan 9 = (a cos B — b sin A)/(b cos J. + a sin 5), 
or 6 = tan- 1 [(a cos B — b sin A) /(b cos A + a sin B)]. 

Squaring (1) and (2) and adding, we obtain 

r 2 = [a 2 + 6 2 - 2 a& (sin ^4 cos B - cos A sin B)] /cos 2 (A - B) 
.-. r = Va 2 + o 2 - 2 ab sin (A - B)/cos {A-B). 



138 PLANE TRIGONOMETRY 

107. Solve the system 

cos (x + y) + cos (x - y) = 2, (1) 1 

sin (x/2) + sin (y/2) = 0, (2) j W 

for values of x and y less than 2 7T. 

By [8] and [10] , from (1) we obtain the equivalent equation 

cos x cos y — sin x sin y + cos x cos y + sin x sin y = 2, 
or cos x cos?/ = 1. (3) 

Equation (3) is satisfied only when cos x and cos ?/ are both + 1 or both 
— 1. Hence x and y are both coterminal with or both coterminal with it. 
Erom (2), sin (x/2) and sin (y/2) must be opposite in quality or both be zero. 

Hence x = 0, y = 0; x ■= ft, y — — it\ x — — it, y — it. 

Observe that either x = 7T, y — it, or x = — 7T, y = — it is a solution of 
(3), but neither is a solution of (2). 

108. Solve for R and F the system 

W - F sin h - R cos h = 0, "1 
TT - Fcos h - R sin A = 0. J 

Observe that this system is algebraic and linear in R and F. 

.: R = Tf/(sin h + cos ft) = F. 

109. Eliminate from the system 

x = r(0-sin0), (1) "I 

y = r(l-cosd). (2) j 

From (2), y = rvers0, or = vers-^y/r). (3) 

From (2), cos 6 = (r — y)/r. .-. sin = ± V 2 ry — y-/r. (4) 

From (1), (3), (4), x = r vers- 1 (y/r) T ^2 ry - y 2 . (5) 

110. Eliminate from the system 

a cos + 6 sin = c, (1) "| 
d cos + e sin = /. (2) j 

Solving the system for cos and sin 0, we obtain 

. af — cd ce — bf 



ae — bd ae — bd 

Squaring the members of equations (3) and adding, we obtain 
(af-cd) 2 + (ce-bf) 2 = ^ , + CQs2 , = L 
(ae - bd) 2 

.-. (ae - bd) 2 = (af - cd) 2 + (ce - bf) 2 . 



(3) 



EXERCISE XXXII 139 

111. Eliminate 6 from the system 

a cos 8 + b sin 8 = c, "1 
& cos 8 — a sin 8 = d. j 

Solving the system for sin 8 and cos 0, we obtain 

sin d = (be- ad) /(a 2 + & 2 ), 
cos 6 = (bd + ac)/(a 2 + b 2 ). 

Squaring and adding, we obtain 

1 = [(be - ad) 2 + (bd + ac) 2 ]/(a 2 + b 2 ) 2 
• = (c 2 + cZ 2 )/(a 2 + b 2 ). 
.-. a 2 + & 2 = c 2 + d 2 . 

112. Solve the cubic equation x s — Spx + q = 0. (1) 
Putting x = z/ji, we obtain 

z 3 - 3.pn 2 z + qn* = 0. (2) 

By (5) in example 53, we have the identity 

cos 3 A = 4 cos 3 A — 3 cos A. 
or cos 3 A - (3/4) cos ^4 - cos (3 A )/4 = 0. (3) 

Comparing identity (3) with equation (2), we see that cos A is a root of 
(2) when n and A satisfy the conditions 

3pn 2 = 3/4, 
and gn 3 = — cos (3 A)/4. 

Hence n = 1/(2 Vp), 

and cos 3^1 = - 4qn s = - q/(2p*/ 2 ). (4) 

Observe that (4) can be solved trigonometrically as above when p is 
positive and q/(2p Z/2 ) is arithmetically equal to or less than 1. 

If Ai is the principal value of A which satisfies (4), then the values 
A\ + 2 7T/3 and A\ 4- 4 7T/3 also satisfy it. 

Hence the roots of equation (1) are 

cos-4i/n, cos (^4i + 2 ?r/3)/n, and cos (^j. + 4 7t/3)/n, 
i.e. 2 VP cos ^Lx, 2 VP cos (^4i + 2 7T/3), and 2 V.P °os (^ii + 4 ar/3). 

By Algebra, we know that the general cubic equation 

y 3 + 3 ay 2 + by + c = 

can be transformed into one of the type (1) by putting y = x — a. 



140 PLANE TKIGONOMETRY 

113. Solve x 3 + 6 x 2 + 9x + 3 = 0. 

Putting x = y — 2, we obtain y z — 3 y + 1 = 0. 
Putting ?/ = z/ft, we obtain z 3 — 3 ft 2 z -f n 3 = 0. 
Now cos 3 A - (3/4) cos A - (1/4) cos 3 A = 0. 

Hence z = cosA when ft 2 =1/4 and n 3 = — cos(3^.)/4, ie. when n= 1/2, 
and cos 3 A = - 1/2 = cos 120°. 

.-. 3^ii = 120 o , or ^li = 40 o . 
Hence z = cos 40°, cos (40° + 120°), or cos (40° + 240°). 

:. y - 2 cos 40°, 2 cos 160°, or 2 cos 280°. 
.-. x = - 2 + 2 cos40°, - 2 - 2 cos20°, or -2 + 2 cos80°. 

Having given sin 15° = ( V3 - l)/(2 V2), and cos 15° = ( y/S + l)/(2 V2), 
solve each of the following five examples : 

114. Solve x 3 - 24 x - 32 = 0. (1) 

Substituting z/n for x, we obtain 

z 3 - 24 ft 2 x - 32 ft 3 = 0, (2) 

cos 3 ^. - (3/4) cos A - (1/4) cos 3 A = 0. 

.-. 24 rfi = 3/4, or n ^ V2/8, 
i cos 3 A = 32 ft 3 = 32 ( V2/8) 3 . 
.-. cos 3^1 =V2/2. 
J. = 15°. 

Since the roots of (2) are 1/n as large as those of (1), we obtain for the 
three roots of (1) 

^^- O = 2 + 2V3, C ^ 8 ^ = -4,^ 0s255 ! = 2-2V3. 
V2/8 v V2/8 V2/8 

115. Solve x 3 - 6 x 2 + 6 x + 8 = 0. (1) 
Put y + 2 for x in (1), ?/ 3 - 6 y + 4 = 0. (2) 
Substitute z/n for ?/ in (2) , z 3 - 6 ?i 2 x + 4 n 3 = 0. (3) 

cos 3 vl - (3/4) cos A - (1/4) cos 3 A = 0. 
.-. 6 ft 2 = 3/4, or n = V2/4. 
(1/4) cos 3 A = - 4 ft 3 , or cos 3 A = - V2/2. 
.-. A = 45°. 

2/ = cos 45°/ft, cos 165°/?z, cos 285°/ft. 
.-. x = 2 + cos45°/n, 2 - cosl5°/n, 2 + sinl5°/ft. 
.-. x = 4, 1 - V3, 1 + V3. 



EXERCISE XXXII 141 

116. Solve x 3 - 3 x/2 - 1/2 = 0. 

Substitute z/n for x, z 3 - 3 n 2 x/2 - n s /2 = 0. 
cos 3 A - (3/4) cos A - \ cos (3 A) = 0. 

.-. 3 ?i 2 /2 = 3/4, or n = V2/2. 
.-. i cos (3 4) = ?i 3 /2, or cos 3 A = V2/2. 
.•'. ^i = 15°. 
.-. x = cosl5°/w, cosl35 /^, cos255°/w. 
.-. x = (1 + V3)/2, - 1, (1 - V3)/2. 

117. Solve x 3 + 3 x 2 - 1 = 0. (1) 

Substituting y — 1 for x in (1), we obtain 

lf*__3y + l = 0. (2) 

Substituting z/n foxy in (2), we obtain 

z 3 - 3 n 2 y + ?i 3 = 0. (3) 

cos 3 A - (3/4) cos A - (1/4) cos (3 A) = 0. 

.-. 3 n 2 = 3/4, or n = 1/2. 
cos (3 A)/4 = - n 3 . 
.-. A = 40°. 
.-. y = cos40°/?i, cosl60°/?z, cos280°/w. 
.-. x = - 1 + cos40°/n, - 1 - cos20°/n, - 1 + cos280°/?z. 
.-. x = - 1 + 2 cos 40°, - 1 - 2 cos 20°, -1 + 2 cos 280°. 

118. Solve x 3 + 4 x 2 + 2 x - 1 = 0. (1) 

Substituting y — 4/3 for x in (1), we obtain 

?/ 3 - (10/3) y + 29/27 = 0. (2) 

Substitute z/n for y in (2), 

z 3 - (10/3) ?i 2 z + (29/27) n 3 = 0. 
.-. (10/3) >i 2 = 3/4. 

n = 3 V10/20. 
cos(3^1)/4 = -(29/27)n 3 . 

cos (3 A) = - 4 (29/27) (3 V10/20) 3 . 
log [cos (3 A)] = 9.66137 - 10. 
.-. 3 ^4 = 117° 17' 33". 
.-. A-l = 39° 5' 51". 

Rootsare-4/3 + (2V10/3)cos^,wliere^. = 39°5 / 51 // , 159° 5' 51", or 
279° 5" 51". 



142 PLANE TRIGONOMETRY 

EXERCISE XXXIII 
Triangles 

1. Two towers are 3 mi. apart on a plain. The angle of depression of 
one, from a balloon directly above the other, is observed to be 8° 15'. 
How high is the balloon ? 

Let x = the height of the balloon in miles. 

Then z = 3 tan 8° 15'. 

log 3 = 0.47712 
log tan 8° 15' = 9.16135-10 
.-. logx = 9.63847 - 10 
.-. z = 0.43498. 

2. The shadow of a tree 101.3 ft. high is found to be 131.5 ft. long. 
Find the elevation of the sun. 

Let A = the elevation of the sun. 

Then * tan A = .101.3/131.5. 

log 101.3= 12.00561 - 10 
log 131.5= 2.11893 
.-. log tan A = 9.88668 - 10 
.:. A = 37° 36' 30". 

3. A rock on the bank of a river is 130 ft. above the water level. From 
a point just opposite the rock on the other bank of the river the angle of 
elevation of the rock is 14° 30 / 21". Find the width of the river. 

Let z = the width of the river in feet. 

Then z = 130 cot 14° 30' 21". 

log 130 = 2.11394 

log cot 14° 30' 21" = 0.58716 

.-. logx = 2.70110 

.-. z = 502.46. 

4. A rope 38 ft. long, when fastened to the top of a tree 29 ft. high, 
just reaches a point in the plane of the foot of the tree. Find the angle 
which the rope makes with the ground. 

Let A be the angle the rope makes with the ground. 
Then sin A = 29/38. 

log 29 = 11.46240-10 

log 38 = 1.57978 
.-. log sin A = 9.88262 - 10 
.-. A = 49° 44' 38". 



EXERCISE XXXIII 143 

5. A window in a house is 24 ft. from the ground. Find the inclina- 
tion of a ladder placed 8 ft. from the side of the building and reaching 
the window. 

Let A = the inclination of the ladder. 

Then tan A = 24/8 = 3. 

.-. A = 71° 33' 54". 

6. A ladder 40 ft. long reaches a window 33 ft. high, on one side of a 
street. Its foot being at the same point, it will reach a window 21 ft. 
high on the opposite side of the street. Find the width of the street. 

Let A and B be the inclinations of the ladder in its two positions. 
Let x and y be the projections of the ladder in its two positions on the 
horizontal plane ; then x + y = the width of the street. 



x=V40 2 -33 2 = V73x7, y = V40 2 - 21 2 = Vei x 19. 

log 73 = 1.86332 ■ log 61 = 1.78533 

log 7 = 0.84510 log 19 = 1.27875 

.-. log x = 2.70842/2 .-. log y = 3.06408/2 

= 1.35421 = 1.53204 

.-. x = 22.605. .-. y = 34.044. 

.-. x + y = 56.649. 

7. A lighthouse 54 ft. high is situated on a rock. The angle of eleva- 
tion of the top of the lighthouse, as observed from a ship, is 4° 52', and 
the angle of elevation of the top of the rock is 4° 2'. Find the height of 
the rock and its distance from the ship. 

Let y = the distance of the rock from the ship, and x = the height of 
the rock ; then x + 54 = the height of the top of the lighthouse. 

tan 4° 52' = (x + 54)/y, cot 4° 2' = y/x. 
.-. x = 54/ (tan 4° 52' cot 4° 2' - 1). 

log tan 4° 52' = 8. 93016 - 10 log 54 = 1 . 73239 

log cot 4° 2' = 1.15174 log . 2075 = 1.31702 

.-. log product = 0.08190 .-. log x = 2.41537 

.-. product - 1 = 1.2075 - 1 = .2075 .-. x = 260.24. 

logx = 2.41537 

log cot 4° 2' = 1.15174 

.-. logy = 3.56711 

.-. y = 3690.7. 



144 PLANE TRIGONOMETRY 

8. A man standing south of a tower, on the same horizontal plane, 

observes its angle of elevation to be 54° 16' ; he goes east 100 yd. , and 

then finds its angle of elevation to be 50° 8'. Eind the height of the tower. 

Let A and C be the two points of observation, and B the foot of the 

tower. Let x be the height of the tower. 

Then CB 2 - AB 2 = AC 2 = 100 2 , 

CB/x = cot 50° 8', 
_ . AB/x = cot 54° 16'. 
.-. CB 2 - AB 2 = 100 2 = (cot 2 50° 8' - cot 2 54° 16') x 2 . 
.-. x = 100/ Vcot 2 50° 8' - cot 2 54° 16'. 
log cot 2 50° 8' = 9. 84352 - 10 log cot 2 54° 16' = 9. 7 1400 - 10 

.-. cot 2 50° 8' = 0.69747. .-. cot 2 54° 16' = 0.51761 + . 



.-. Vcot 2 50° 8' - cot 2 54° 16' = V. 17986 
log 100 = 2.00000 
log V717986 = 9.62746 - 10 
.-. logo; = 2.37254 
.-. x = 235.8. 

9. A pole is fixed on the top of a mound, and the angles of elevation 
of the top and the bottom of the pole are 60° and 30° respectively. 
Prove that the length of the pole is twice the height of the mound. 

Let A be the point of observation, B the foot of the mound, x the 
height of the mound, and y the height of the pole. 

Then AB/x = cot 30° = tan 60°, 

and (z + y)/AB = tan 60°. 

.-. y = (tan 2 60° - l)x = (3 - l)x = 2x. 

10. Given that the radius of the earth is 3963 mi. , and that it subtends an 
angle of 57' 2" at the moon. Find the distance of the moon from the earth. 

Let x = the distance of the moon from the earth. 

Then x = 3963/sin 57' 2". 

log 3963 = 13.59802 -10 
log sin 57' 2" = 8.21983-10 
.-. logx= 5.37819 
.-. x = 238880. 

11. Given that the radius of the earth is 3963 mi. , and that it subtends 
an angle of 9" at the sun. Eind the distance of the sun from the earth. 

Let x = the distance of the sun from the earth. 

Then x = 3963/sin 9". 



EXERCISE XXXIII 



145 



log 3963= 13.59802 -10 
log sin 9" = 5.63982 - 10 
.-. logx = 7.95820 
.-. x = 90824000. 

12. Solve example 1 in § 57 by the principles of right triangles 

The given parts are a side and two angles. 

Given a = 180, A = 38°, B = 75° 13'; to find 6, C, c. 

Draw CH _L AB. 

C = 180° - (A + B) = 66° 17'. 
BH = a cos B. 



HA = CHcotA. 

c = BH + HA. 

log a = 2.25527 

log cos B = 9.39220 - 10 

•. log £#= 1.64747 

.-. BH = 44.409. 

log CH= 2.24163 

log cot A = 0.10719 



CH=a sin B. 

b = CH/sinA. 

loga= 2.25527 

log sin B = 9.98636 -10 

-. log CH= 12.24163 -10 

log sin A = 9.78934 -10 

.-. b = 2.45229 

b = 283.33. 



.-. log HA = 2.34882 
.-. HA = 223.26. 

.-. c = BH + HA = 267.67. 

13. Solve the first four examples in Exercise XIX by the principles of 
right triangles. 

(1) Given B = 60° 15', C = 54° 30', a = 100 ; to find A, 6, c. 
Draw CH±AB. c = AH + HB. 

A = 180° - (A + B) = 65° 15'. 



CH = a sin B. 
b= CH /sin A. 
loga= 2.000 
log sin B = 9.93862 - 10 
•. log CH= 11.93862 -10 
log sin A = 9.95815 - 10 
.-. log 6= 1.98047 
.-. b = 95.6025. 



HB = a cos B. 
AH= CH cot A. 
log a = 2.00000 
log cos B = 9.69567 - 10 
■. logHB= 1.69567 
.-. HB = 49.621. 

log Cfl"= 1.93862 
log cot A = 9.66371 - 10 
. log^£T= 1.60233 

.-. AH= 40.025. 
.-. c = 89.646. 



146 



PLANE TRIGONOMETRY 



(2) Given A = 45° 41', C = 62° 5', b = 
Draw AH" JL JSC. 

J5 = 180° -(A+C) = 72° 14' 
^.jEZ" = b sin C. 
c = A B = ^iff/sin 5. 
.-. yl/i" = 100 sin 62° 5'= 88.36. 
logoff =1.94626 
log sin B = 9.97878 - 10 
.-. logc = 1.96748 
.-. c = 92.786. 



(3) Given B 

A = 

Draw BH± 

CH = 

c = 

log a = 

log cos C = 

.-. log OH = 

.-. CH = 

log ff£ = 

log sin A = 

.-. log c = 

.-. c = 

(4) Given ^L 
Draw A H ± 

C = 

. AH = 

b = 

logc = 

log sin B = 

.-. log AH = 

log sin C = 

.-. logb = 

.-. b = 



= 70° 30', C = 78° 10', a 
180° - (J5 + C) = 31° 20' 
4C. 
a cos C. 
HjB/sin ^4. 
2.00860 
9.31189 -10 
1.32049 
20.917. 

11.99927 - 10 
9.71602 - 10 



2.28325 
191.98. 

= 55°, B = 65°, c = 270 ; to find a, 6, C, 
BC. 

180° -(A + B) = 60°. 
c sin B. 
AH/sin C. 

2.43136 

9.95728 - 10 



100 ; to find a, B, c. 



HC = b cos G. 
BH = AH cot B. 
.-. ff C = 100 cos 65° 5'= 46.82. 
log .Iff = 1.94626 
log cot B = 9.50572 -10 
■. log BH= 1.45198 
.-. BH = 28.312. 

.-. a = BH + HC = 75.132. 
= 102 ; to find B, 6, c. 



HB = a sin C. 

ffA = ff£cot.l. 

log a = 2.00860 

log sin C = 9.99067 - 10 

. log HB= 1.99927 

log cot A = 0.21552 

. log HA = 2.21477 

.-. ff4 =163.97. 

.-. 6 = CH+HA = 184.89. 



12.38864 
9.93753 



2.45111 
282.56. 



£ff = c cos 5. 
HG = AH cot G. 

logc = 2.43136 

log cos B = 9.62595 - 10 

\ logoff =2.05731 

.-. BH= 114.10. 

logoff =2.38864 

log cot C = 9.76144 -10 

.-. log Cff = 2.15008 

.-. ffC = 141.28. 

.-. a = BH + HC = 



255.38. 



EXEliCiSE XXXIII 



147 



14. Solve example 1 in § 58 by the principles of right triangles. 
Given a = 250, b = 240, A = 72° 4' ; to find B, C, c. 
Draw CH±AB. 



AH = bcosA. 




CII = b sin A. 




HB = HC cot B. 




sin 5 = CR/a. 




log 6 = 2.38021 




log 6= 2.38021 




log cos A =9.48842 - 


10 


log sin A = 9.97837- 
.-. log CH = 12.35858 - 


10 


. log AIT =1.86863 




10 


.-. All =73.902. 




loga= 2.39794 




log CH= 2.35858 




.-. log sin £ = 9.96064 - 


10 


log cot 5 = 9.64912 - 


10 


.-. B = 65° 58' 24" 
.-. C = 41° 57' 36" 




•. log B1I = 2.00770 






.-. BH = 101.79. 




.-. c=AH + BH 
= 175.69. 





15. Solve examples 1, 3, 5, and 7 in Exercise XX by the principles of 
right triangles. 

5 = 41° 10'; to find A, C, c. 



(1) Given a = 145, b = 17! . 
Draw CH±AB. 

CH=a sin B. 
sinJ.= CH/b. 
loga= 2.16137 
log sin B = 9.81839 -10 
.-. log CH= 11.97976 -10 
log 6= 2.25042 
.-. log sin A = 9.72934 -10 
.-. A = 32° 25' 36". 
.-. C = 106° 24' 24". 



in? = a cos jB. 

C = 180° - (4 + £). 
AH = CH cot A. 
log a = 2.16137 
log cos 5 = 9.87668 -10 
\ log JB J S'= 2.03805 
.-. BIT =109.16. 
log CH= 1.97976 
log cot A = 0.19704 
. logoff = 2.17680 
.-. AH= 150.24. 
.-. c=AH+HB 
= 259.4. 



148 



PLANE TRIGONOMETRY 



(3) Given a = 5.98, b = 3.59, A = 63° 50'; to find B, C, c. 
Draw CH±AB. 



CH=b sin A. 


AH = &cos A. 


sin B = CH/a. 


HB= CH cot B. 


log& = 0.55509 


log 6 = 0.55509 


log sin A= 9.95304-10 


log cos A = 9.64442 -10 


.-. log CH = 10.50813 - 10 


.-. logoff =0.19951 


loga = 0.77670 


.-. AH= 1.583. 


.-. log sin B= 9.73143-10 


log CH= 0.50813 


.-. B = 32° 36' 9". 


log cot 5 = 0.19411 


.-. C = 83° 33' 51". 


.-. log HB = 0.70224 




.-. HB= 5.038. 


- 


.-. c = 6.621. 


(5) Given 6 = 74.1, c = 64.2, C = 27° 18'; 


to find A, a, 5. 


Draw AH± BC or BC produced. 




AH =b sin C. 


C.ff =b cos C. 


sin B = AH/c. 


£T£ = c cos 5. 


log 6= 1.86982 


log 6= 1.86982 


log sin C = 9.66148-10 


log cos C = 9.94871 -10 



. log AH = 11.53130 -10 

logc= 1.80754 ' 
log sin £ = 9.72376-10 
.-. Bi = 31° 57' 46 // . 



.-. log CH= 1.81853 
.-. CJJ=65.846. 
logc = 1.80754 
log cos 5 = 9.92860 - 10 



B 2 = 148° 2' 14". 
A 1 = 120° 44' 14' 
A 2 = 4° 39' 46". 



.-. log BH= 1.73614 
.-. HB = 54.4675. 
.-. «! = Off + -ffB 
= 120.31. 
a 2 = CH- HB 
= 11.379. 
Since c<b and log c > log AH, we know that there are two solutions. 



EXERCISE XXXIII 



149 



(7) Given b = 45.21, c = 50.3, B = 


40° 32' T'\ to find A, a, C. 


Draw AH± BC or BC produced. 




AH = c sin B. 


BH = c cos B. 


sin C = AH/b. 


HC = AHcotC. 


logc = 1.70157 


logc= 1.70157 


log sin B= 9.81286-10 


log cos B = 9.88082 -10 


.-. log AH = 11.51443-10 


.-. log BH= 1.58239 


log 6= 1.65523 


.-. BH= 38.23 


.-. log sin C = 9.85920-10 


log AH = 1.51443 


.-. Ci = 46° 18' 40". 


log cot (7 = 9.98015 -10 


C. z = 133° 41' 20". 


.-. log HC = 1.49458 


.-. A x = 93° 9' 13". 


.-. HC= 31.23. 


A 2 = 5° 46' 33". 


.-. a 1 = Sfi" + HC 




= 69.46. 




a 2 = -frff - B"C = 7 



16. Solve the example in § 59 by the principles of right triangles. 
The given parts are two sides and their included angle. 
Given a = 540, b = 420, C = 52° 6'; to find A, B, c. 
Draw BH ± CA. 



CH = a cos C. 
c = AH/cos A. 
log a = 2.73239 
log cos C = 9.78837-10 
.-. log CH = 2.52076 
.-. OH" = 331.71. 
AH =b-CH= 88.29. 



log AH = 1.94591 

log cos A = 9.30727 

.-. logc 



10 



2.63864 
= 435.15. 



BH = a sin C. 
tan J. = BH/AH. 
log a = 2.73239 
log sin = 9.89712 -10 
.-. log BH= 2.62951 
log AH= 1.94591 
log tan J. = 0.68360 
.-. A = 78° 17' 37". 
.-. B = 49° 36' 23". 



150 



PLANE TRIGONOMETRY 



17. Solve the first four examples in Exercise XXI by the principles of 
right triangles. 



(1) Given a = 266, b = 352, C = 73°; 


to find .4, £, c. 


Draw BH±CA. 


B = 180° -(A + C). 




BH = a sin C. 


CJT = a cos C. 


tan A = BH/AH. 


c = AH /cos A. 


log a = 2.42488 


log a = 2,42488 


log sin C = 9.98060 -10 


log cos C = 9.46594 -10 


.-. log BH= 12.40548 -10 


.-. log CH= 1.89082 


log AH = -2.43811 


.-. Cil= 77.77 + . 


.-. log tan A = 9.96737 -10 


. AH =+b-CH -274.23. 


.-. A = 42°. 50' 56 ,/ . 


log J.JJ = 12.43811 - 10 


.-. B = 64° 9' 4". 


logcosJL = 9.86519-10 




.-. logc = 2.57292 


. 


.-. c = 374.05. 


(2) Given b = 91.7, c = 31.2, A = 33 c 


' V 9"; to find a, 5, C. 


Draw BR ± AC. 




BH — c sin A. 


5 = 180° -(A-+ C). 


tan C = HB/HC. 


AH = ccos J.. 


logc = 1.49415 


a = BH/sin C. 


log sin A= 9.73750-10 


logc = 1.49415 


.-. \ogBH= 11.23165 -10 


log cos ^4 = 9.92301 -10 


logHC = 1.81670 


.-. log AH = 1.41716 


.-. log tan C = 9.41495-10 


.:. AH" =26.131. 


.-. C = 14° 34' 24". 


.:HC = b-AH = 65.569 


.-. B = 132° 18 / 27 // . 


log BH= 11.23165 -10 




log sin C= 9.40074-10 




.-. loga= 1.83091 




.-. a = 67.75. 



EXEECISE XXXIII 



151 



(3) Given a = 960, b = 720, C = 25° 40'; to find A, B, c. 
Draw AH±BC. 

AH = b sin C. 

HC = b cos C. 

log 5 = 2.85733 

log sin C = 9.63662 -10 



. log ^.#=2.49395 
log 6 = 2.85733 
log cos C = 9.95488 -10 
. ^#0 = 2.81221 

.-. HC = 648.95. 

.-. BH= a- H C = S 11.05. 



tanJ3 : 

C : 

log AH ■- 
log BH : 

. log tan B - 
.: B-- 

= 180 -(B 
log BH -. 

log COS B : 
.-. logC: 



AH/BH 
BH/cos B. 
2.49395 

2.49283 



(4) Given a = 886, b = 747, C = 71° 54'; to find A, 
Draw BH _LCA. 

BH = a sin C. CH 

tan J. = BH/AH c 

\oga = 2.94743 

log sin C = 9.97796 -10 



.-. logoff = 2.92539 
log AH= 2.67371 

log tan A = 0.25168 
.-. A = 60° 44' 38' 
.-. B = 47° 21' 22' 



log a 

log cos C 

\ log Oil 

.-. OH" 

.-. AH 

log AH ■ 

log cos A ■■ 

.-. lOgC: 

.-. c 



= 0.00112 
= 45° 4' 26 // . 
■f C) = 109° 15' 34' 
= 12.49283 - 10 
= 9.84892 - 10 
= 2.64391 
= 440.46. 
B, c. 

= a cos C. 
= AH /cos A. 
= 2.94743 
= 9.49231 - 10 
= 2.43974 
= 275.256. 
,471.744. 
= 12.67371 -10 
= 9.68906-10 
= 2.98465 
= 965.28. 



18. Solve example 6 in § 60 by the principles of right triangles. 
The given parts are the three sides. 

In the triangle ABC draw CH ± AB ; let AH = y, HB = x. 
Then 6 2 _ y i - cm = a 2 - x 2 . 

.-. x 2 - y 2 = a 2 - 6 2 = (a + 6) (a - 6). 
.-. x — y = (a + b) (a — 6)/c, since x 4- y = c. 



152 



PLANE TRIGONOMETRY 



Given a = 130, b = 123, c = 77; to 
Here a + b = 253, a - b = 7. 
log(a + &)= 2.40312 
log (a - b) = 0.84510 
.*. log product = 3.24822 
log c = 1.88649 
.-. log(x-2/) = 1.36173 
.-. x-y = 23. 
cc + y = 77. 
.-. x = 50. 
2/ = 27. 



find A, B,C. 

cos J. = ?//&. 

cos B = x/a. 

logy = 11.43136-10 

log b = 2.08991 

.-. log cos A = 9.34145-10 
.-. A = 77° 19' 12". 
logx = 11.69897 -10 
loga = 2.11394 

.-. log cos B= 9.58503 -10 



19. Solve the 
of right triangles 

(1) Given a = 

x - y 

Here a + b 

log{a + b) 

log(a-6) 

.-. log product 

logc 

.-. log{x-y) 

.-. x - y 

x + y 

.-. x 

y 

(2) Given a = ! 

y -x 

Here 6 + a 

log (b + a) 

log (6 - a) 

.-. log product 

logc 

.-. log (y - x) 

••• y - x 
y + x 

••• y 

X 



.-. B = 67° 22' 48". 
C = 180 ^ (A + E) = 35° 18'. 

first four examples in Exercise XXII by the principles 



56, b = 43, c = 49 
= (a'+b)(a-b)/c. 
= 99, a - 6 = 13. 
= 1.99564 
= 1.11394 



to find A, B, C. 

cos A = 
cos B = 
logy = 
log 6 = 



y/b. 

x/a. 

11.05565 - 10 
1.63347 



= 3.10958 


.-. log cos A= 9.42218-10 


= 1.69020 


.-. A = 70° 40' 18". 


= 1.41938 


logo; = 11.57557 -10 


= 26.265. 


loga= 1.74819 


= 49. 


.-. logcosJB= 9.82738 -10 


= 37.633. 


.-. B = 47° 46' 39". 


= 11.367. 


C = 180° - (A + B) = 61° 33' 3' 


8.5, 6 = 9.2. c = 7.8 


to find A, B, C. 


= (b+a)(b-a)/c. 


cos A = y/b. 


= 17.7, b-a = .7. 


cos B = x/a. 


= 1.24797 


logy = 10.67157 -10 


= 1.84510 


log&= 0.96379 


= 1.09307 


.-. log cos A = 9.70778 - 10 


= 0.89209 


.-. A = 59° 19' 11". 


= 0.20098 


logx= 10.49216-10 


= 1.5885. 


log a = 0.92942 



= 7.8. 
= 4.6943. 
= 3.1057. 



.-. log cos B 

.-. B = 
C = 180° - (A 



9.56274 - 10 
68° 34' 9". 
+ £) = 52° 6' 40' 



EXERCISE XXXIII 



153 



(3) Given a = 61.3, b = 84.7, c = 47.1 
Draw BH± AC ; let AH = y, HC = 

Then x — y = (a + c)(a — c)/6. 
Here a + c = 108.9, a-c = 13.7. 

log(a + c) = 2.03703 
log(a-c) = 1.13672 
.*. log product = 3.17375 
log b = 1.92788 
.-. log(x-y)= 1.24587 
.-. x - y = 17.614. 
x + y = 84.7. 
.-. a: = 51.157. 
y = 33.543. - 



to find A,% C. 

cos A = y/c. 
cos C = x/a. 

logy = 11.52560 -10 

log c = 1.67761 
.-. log cos A = 9.84799-10 

.-. A = 45° ir 46 y/ . 

logx = 11.70891 -10 



lOi 



log a 
cos C 
.: C 



1.78746 



10 



B = 180 c 



9.92145 
33° 25' 52". 
(vl + C) = 101° 22' 22 ,/ . 



(4) Given a = 705, b = 562, c = 639 ; to find A, B, C. 



x — y = (a + b)(a — b)/c. 
Here a + 6 = 1267, a - & = 143. 

log(a + b) = 3.10278 
log (a- b) = 2.15534 
.-. log product = 5.25812 
log c = 2.80550 
.-. log(x - y) - 2.45262 
.-. x-y= 283.54. 
x + y = 639. 
.-. x = 461.27. 
y = 177.73. 



cos ^1 = y/6. 

cos B — x/a. 

logy = 12.24976 -10 

log 5= 2.74974 

. log cos A = 9.50002 -10 
.-. A = 71° 33' 51 // . 
log x = 12.66395 -10 
log a = 2.84819 

■. logcosJ5= 9.81576 - 10 
.-. B = 49° 8' 8". 

180° - (A + B) = 59° 18' l /y 



20. A tree stands at a distance from a straight road and between two 
milestones. At one milestone the line to the tree is observed to make an 
angle of 25° 15' with the road, and at the other an angle of 45° 17'. Find 
the distance of the tree from the road. 

Let A and C be the milestones and B the tree. 
Vr&v? BII±AC. 

cot 25° 15' = AH/HB. 
cot 45° 17'= RC/HB. 
.-. cot 25° 15' + cot 45° 17' = {AH + HC)/HB = 1/HB. 
.-. HB = l/(cot 25° 15' + cot 45° 17 r ) 
= 1/(2. 1203 + .9902) 
= .32149. 



154 PLANE TRIGONOMETRY 

21. From the decks of two ships at C and D, 880 yd. apart, a cloud A, 
in the same vertical plane as C and D and between them, is observed. Its 
angle of elevation at C is found to be 35°, and at D 64°. Find the height 
of the cloud above the surface of the sea, the height of the eye in each 
case being 21 ft. 

Draw AHA. CD. 

cot 35° = CH/AH. 

cot 64° = HD/AH. 

■/. cot 35° + cot 64° = 880/ AH. 

.-. AH = 880/(cot 35° + cot 64°) 
= 880/(1.4281 + 0.4877) 
= 459.34. 
AH+ 7 =466.34. 

22. To determine the distance between two ships at sea, an observer 
noted the interval between the flash and report of a gun fired on board 
each ship, and measured the angle which the two ships subtended. The 
intervals were 4 seconds and 6 seconds respectively, and the angle 48° 42'. 
Find the distance between the ships, the velocity of sound being 1142 ft. 
per second. 

Let A and B be the two ships and C the observer. 

Given 6 = 4 x 1142 = 4568, a = 6x 1142 = 6852, C = 48° 42'; to find c. 

A-B o-6 C 

tan = cot — • 

2 a + b 2 

(A + B)/2 = 90° - C/2 = 65° 39'. 
c = a sin C/sin A. 
log (a- b) = 13.37731 -10 loga= 3.83582 

log (a + 6) = 4.05767 log sin C = 9.87579-10 

.-. log quotient = 9.31964-10 .-. log product = 13.71161 - 10 

log cot (C/2) = 0.34432 log sin A = 9.99999 - 10 

.: log tan i {A-B) = 9.66396-10 .-. log c = 3.71162 

.-. (A - B)/2 = 24° 45' 45 // . .-. c = 5147.9. 

.-. A = 90° 24 r 45". 

23. In order to find the breadth of a river a base line of 500 yd. was 
measured in a straight line close to one side of it, and at each extremity 
of the base the angle subtended by the other end and a tree upon the 
opposite bank was measured. These angles were 53° and 79° 12' respec- 
tively. Find the breadth of the river. 

Call the extremities of the measured line A and 5, and the tree C. 
Draw CH±AB. 



EXERCISE XXXIII 155 

cot 53° = AH/HC. 

cot 79° 12' = HB/HC. 

.-. cot 53° + cot 79° 12' = 500/ EC. 

.-. HC = 500/(cot 53° + cot 79° 12') 
= 500/ (.7536 + .1908) 
= 529.4. 

24. A straight road leads from a town A to a town B, 12 mi. distant ; 
another road, making an angle of 77° with the first, goes from A to a 



town C, 7 mi. distant. Find the distance between the towns B and C. 


Given A = 77°, b = 7, c = 12 ; to find 


a. 


tan = cot — • 

2 c + b 2 


(C + B)/2 = 90° 


- ^4/2 = 51° 30'. 


a = c sir 


i A/s'm C. 


log(c-6) = 10.69897 -10 


logc= 1.07918 


log(c + 6)= 1.27875 


log sin A= 9.98872 


.-. log quotient = 9.42022 - 10 


.-. log product = 11.06790 - 10 


log cot {A/2)= 0.09939 


log sin C= 9.97245-10 


.-. log tan £((7-5)= 9.51961-10 


.-. logct= 1.09545 


.-. (C - B)/2 = 18° 18' 21". 


.-. a = 12.458. 


(C + £)/2 = 51° 30'. 




.-. C = 69° 48' 21". 





25. Two lighthouses A and B are 11 mi. apart. A ship C is observed 
from them to make the angles BAC = 31° 13' 31" and ABC = 21° 46' 8". 
Find the distance of the ship from A. 

Given A = 31° 13' 31", B = 21° 46' 8", c = 11 ; to find b. 

C = 180° -{A + B) = 127° 21". 
b = c sin Z?/sin C. 
\ogc= 1.04139 
log sin B = 9.56921 -10 
.-. log product = 10.61060 - 10 
log sin C = 9.90232-10 
.-. log 6= 0.70828 
.-. 6 = 5.1083. 

26. Two posts A and B are separated by a swamp. To find the distance 
between them a point C is so taken that both posts are visible from it. By- 
measurement, A C= 1272.5 ft., BC = 2012.4 ft., and ZACB = 41° 9' 11". 
Find the distance AB. 



156 



PLANE TRIGONOMETRY 



Given a = 2012.4, b 



tan 



1272.5, C-41 
A-B a- 



Y 11"; to find c. 



b C 
— cot — . 
2 a + b 2 



{A + B)/2 = 90° - C/2 = 69° 25' 24| 
c = a sin C/sin J.. 



log (a -6) = 12.86917 -10 
log (a + 6) = 3.51669 
..-. log quotient = 9.35248 - 10 

log cot (C/2) = 0.42550 
log tan \{A -B) = 9.77798-10 
.-. (A - B)/2 = 30° 57' 14". 
.-. J. = 100° 11' 19". 



loga= 3.30371 
log sin C = 9.81827 - 10 
log product = 13.12198 - 10 
log sin A = 9.99284 -10 
.-. logc= 3.12914 
.-. c = 1346.3. 



27. Two buoys A and B are one half mile apart. Find the distance 
from A to a point C on the shore if the angles ABC and BAC are 77° T 
and 67° 17 r respectively. 

Given A = 67° 17', B = 77° 7 r , c .= .5 ; to find 6. 
6 = c sin 5/sin C. 
O = 180° -(A + B) = 35° 36 / . 
logc= 9.69897 -10 
log sin B = 9.98893 - 10 
.-. log product = 19.68790 - 20 
log sin C = 9.76501 -10 
.-. log 6= 9.92289 -10 
.-. b = .83732. 

28. The elevation of the top of a spire at one station, A, was 23° 50' 15", 
and the horizontal angle at this station between- the spire and another 
station, 5, was 93° 4' 15". The horizontal angle at B was 54° 28" 30", and 
the distance between the stations 416 ft. Find the height of the spire. 

Let T be the top of the spire and C the base. 



In A ABC, A 
B 



and 



93° 4' 15", 
54° 28' 30", 
c = 416. 
. C = 32° 27' 15", 
AC = c sin B/s'mC. 
logc= 2.61909 
log sin B = 9.91055 - 10 
log product = 12.52964 - 10 
log sin C = 9.72972 - 10 
.-. log^4C= 2.79992 



In A CTA, 

Z(L4T=23°50 / 15", 

and CT= AC tan CAT. 

log AC = 2.79992 

log tan CAT= 9.64526 - 10 

.-. log CT= 2.44518 

.-. CT= 278.7. 



EXERCISE XXXIII 



157 



29. In order to find the distance of a battery at B from a fort at F, 
distances BA and AC were measured to points A and C from which 
both the fort and the battery were visible, the former distance being 
2000 and the latter 3000 yd. The following angles were then measured : 
Z BAF = 34° 10', ZFAC = 74° 42', and Z FCA = 80° 10'. Find the dis- 
tance of the fort from the battery. 



In A A CF, A = 74° 42', 
C = 80° 10', 

^1C = 3000. 
•. F = 180° - (A + C) = 25° 8'. 

FA = AC sin C/sinF. 

log AC = 3.47712 
log sin C = 9.99357 - 10 
.-. log product = 13.47069 - 10 
log sin F = 9.62811 - 10 
.-. log^i^= 3.84258 

.-. AF= 6959.5. 



In AAFB, ^4 = 34° 10', 
/ = 2000, 
6 = ^4^=6959.5. 
F)/2 = 72° 55'. 
B-F b 



tan 



/ A 
cos — • 

b+f 2 

a = b sin A /sin B. 

6+/= 8959.5. 
b -/=4959.5. 

log (&-/)= 13.69544 -10 
log (&+/)= 3.95229 

.-. log quotient 
log cot {A/2) 

logtani(£-F)= 0.25556 



9.74315 
0.51241 



10 



(B 



F)/2 = 60° 57' 42" 
.-. B = 133° 52' 42' 



log 6= 3.84258 
log sin A = 9.74943 - 10 
log product = 13.59201 - 10 
log sin B = 9.85785 - 10 
.-. loga= 3.73416 
.-. a = 5422. 



30. The distances of two islands from a buoy are 3 and 4 mi. respec- 
tively. The islands are 2 mi. apart. Find the angle subtended by the 
islands at the buoy. 

Let B and C be the islands and A the buoy ; then we have a = 2, 
6 = 4, c = 3; to find A. 

6 2 + c 2 - a 2 _ 21 
~~24 



cos A 
.-. A 



2 be 
28° 57'. 



.8750. 



158 PLANE TRIGONOMETRY 

31. Two rocks in a bay are c yd. apart, and from the top of a cliff in 
the same vertical plane with the rocks their respective angles of depression 
are A and 3 A. Show that the height of the cliff is c sin 3 A/(2 cos A). 

Let B be the farther and C the nearer rock, T the top of the cliff and 
R the point where the vertical line from T cuts the horizontal plane 
through B. 

Then cot A = BH/H T, 

cot 3 A = CH/HT. 
.-. cot A -cot 3 A = {BH- CH)/HT = c/HT. 
.-. HT = c/(cot A-cotZA) 

= c sin A sin 3 A/ [sin 3 A cos A — cos 3 A sin A) 

= c sin A sin 3 A/ sin 2 A 

= csin A sin 3 A/(2 sin A cos J.) 

= csin 3 A/(2cosA). 

32. A person wishes to find the distance between two places A and B 
on opposite sides of a v brook. He walks from B to a bridge 2 mi. away. 
Crossing this he continues his walk 6 mi. in the same direction to C, 
which he knows to be 3 mi. from A, If A is 4 mi. from the bridge, 
show that AB = 5.86 mi., nearly. 

Let H be the bridge ; then in A A CH, AC = 3, AH = 4, GH = 6. 

42 + 6 2 _ 32 43 



cos J.HC 



2-4-6 48 



In A ABH, AB 2 = 4 2 + 2 2 +2 - 4 - 2 cos AHC 

= 34.3333. 
.-. logJ_5 = 1.53571/2 
= 0.76786. 
.-. ^4J5= 5.86 mi., nearly. 

33. A person at the top of a mountain observes the angle of depression 
of an object in the horizontal plane beneath to be 45°; turning through an 
angle of 30° he finds the depression of another object in the plane to be 
30°. Show that the distance between the objects is equal to the height of 
the mountain. 

Let D be the top of the mountain, A the first object, B the second, C 
the foot of the vertical line from I) to the horizontal plane through A 
and B. Let CD = h. 

Then AC=CD = h, 

BC = h cot 30° = h V3, 
ZACB = 30° = C. 
In A ABC, c 2 = a 2 + 6 2 -2a6cosC 

= 3 A 2 + W - 2 V3 A 2 V3/2 = ft 2 , or c=h. 



EXERCISE XXXIII 159 

34. From a window on a level with the bottom of a steeple the angle of 
elevation of the top of the steeple was 40°. At another window 18 ft. 
vertically above the former, the angle of elevation was 37° 30'. Find the 
height of the steeple. 

Let A be the first window, B the second, C the top of the steeple, and R 
the base ; and let x = the height of the steeple, 
tan 37° 30' = (x-lS)/RA^ 
cot 40° = RA/x J 

.-. x = 18/(1 - tan 37° 30' cot 40°). 
log tan 37° 30' = 9.88498 - 10 log 18 = 11.25527 - 10 

log cot 40° = 0.07619 log .085525 = 8.93209 - 10 

.-. log product = 9.96117 - 10 .-. log x = 2.32318 

.-. product = 0.914475 .-. x = 210.46. 

.-. 1 - tan 37° 30' cot 40° = .085525. 

35. Find what angle a tower will subtend at a distance equal to six 
times the height of the tower. Find where an observer must station 
himself that the angle of elevation may be double the former angle. 

Let B be the top and C the base of the tower, A the first point of 
observation and A' the second. 

Then tan A =1/6 = .1667. 

.-. A = 9° 28'. 

A'C = cot 2 A times the height of the tower 
= 2.9152 times the height of the tower. 

36. Two ships are a mile apart. The angular distance of the first ship 
from a fort on the shore, as observed from the second ship, is 35° 14' 10"; 
the angular distance of the second ship from the fort, observed from the 
first ship, is 42° 11' 53". Find the distance in feet from each ship to 
the fort. 

Let A be the first ship, B the second ship, and C the fort. 
Given A = 42° 11' 53", B = 35° 14' 10", c = 5280 ; to find a and 6. 
C = 180° -(A+B)= 102° 33' 57". 
b = c sin B/sin C. 
a = c sin A/sin C. 
logc= 3.72263 log c = 3.72263 

log sin 5= 9.76114 -10 log sin A= 9.82717-10 

.-. log product = 13.48377 - 10 .-. log product = 13.54980 - 10 

log sin (7= 9.98947 -10 log sin C= 9.98947 - 10 

.-. log&= 3.49430 .-. loga= 3.56033 

.-. 6 = 3121.1. .-. a = 3633.5. 



160 PLANE TRIGONOMETRY 

37. The sides of a triangle are 17, 21, 28. Prove that the length of a 
line bisecting the greatest side and drawn from the vertex of the opposite 
angle is 13. 

Let a = 17, b = 21, c = 28, and m = the medial line. 

Then cos B = (17 2 + 28 2 - 21 2 )/2 ■ 17 • 28, 

17 2 + 28 2 - 21 2 

m 2 = 17 2 + 14 2 - 2 • IH ■ H ^ - 

^ ^ 2- #.20 

= 17 2 - 2 • 14 2 + 21 2 = 169 

2 
.-. m = 13. 

38. Along the bank of a river is drawn a base line of 500 ft. The 
angular distance of one end of this line from an object on the opposite 
side of the river, as observed from the other end of the line, is 53°; the 
angular distance of the second extremity from the same object, observed 
from the first extremity, is 79° 12'. Find the breadth of the river. 

For solution, see example 23. 

39. Two forces, one of 410 lb. and the other of 320 lb., makes an angle 
of 51° 37'. Find the size and direction of their resultant. 

Given a = 410, b = 320, C = 128° 23' ; to find c and B. 

A-B a-b C 
tan = cot — • 

2 a + b 2 

(A + B)/2 = 90° - C/2 = 25° 48' 30". 
c = b sin C/sin B, 
log (a - b) = 1.95424 log 6 = 2.50515 

log (a- + b) = 2.86332 log sin C = 9. 89425 - 10 

.-. log quotient == 9.09092 - 10 .-. log product = 12.39940 - 10 

log cot (C/2) = 9.68449-10 log sin B= 9.58094-10 

.-. logtan(J.-2?)/2 = 8.77541 -10 .-. logc= 2.81846 

.-. (A - B)/2 = 3° 24' 43". .-. c = 658.36. 

• .-. B = 22° 23' 47". 



40. An unknown force combined with one of 128 lb. produces a result- 
ant of 200 lb., and this resultant makes an angle of 18° 24' with the known 
force. Find the size and direction of the unknown force. 



EXERCISE XXXIII 161 

Given A = 18° 24', b = 128, c = 200 ; to find a and B. 

C-B c-b A 

tan = ■ cot — 

2 c + b 2 

(C + B)/2 = 90° - 4/2 = 80° 48'. 
a = b sin J. /sin 2?. 
log(c- 6) =11.85733 -10 log& = 2.10721 

log(c + 6) = 2.51587 log sin J. = 9.49920 - 10 

.-. log quotient = 9.34146 - 10 .-. log product = 11.60641 - 10 
log cot (A/2) = 0.79058 log sin B = 9.66032 - 10 

.-. logtan(C-E)/2 = 0.13204 .\loga= 1.94609 

.-. (G - B)/2 = 53° 34' 44". .-. a = 88.326. 

.-. C = 134°22 / 44". 

Hence the unknown force is one of 88.326 lb., and makes an angle of 
45° 37' 16" with the known force of 128 lb. 

41. Two sides of a parallelogram are 59.8 ch. and 37.05 ch., and the 
included angle is 72° 10'. Find the area. 

Here A = 72° 10', b = 59.8, c = 37.05 ; to find area of EJ. 

Area of parallelogram = be sin A. 

log 6 = 1.77670 

logc = 1.56879 

log sin A = 9.97861 -10 

.-. log area = 3.32410 

.-. area = 2109.1 sq. ch. 

= 210 acres 9.1 sq. ch. 

42. The three sides of a triangle are 49 ch., 50.25 ch., and 25.69 ch. 
Find the area. 

Given a = 49, b = 50.25, c = 25.69 ; to find the area of the triangle. 



F = Vs(s- a)(s -b)(s- c). 
s = (a + b + c)/2 = 62.47; s - a = 13.47; s - b = 12.22 ; s - c = 36.78. 
logs= 1.79567 
• log (s -a) = 1.12937 
log(s- b) = 1.08707 
log (s - c) = 1.56561 
.-. \ogF= 5.57772/2 
= 2.78886. 
.-. F = 614.975 sq. ch. 

= 61 acres 4.975 sq. ch. 



162 PLANE TRIGONOMETRY 

43. One side of a regular pentagon is 25. Find the area. 

C/2 = 180°/w = 36°. 

h = (c/2) cot (C/2) = 12. 5 cot 36°. 
p = 5c = 5x 25 = 125. 
F = ph/2 = 125 x 12.5 cot36°/2 
= 62.5 x 12.5 cot 36°. 
log-62.5 = 1.79588 
log 12. 5 = 1.09691 
log cot 36° = 0.13874 
.-. log F= 3.03153 
.'. F= 1075.3. 

44. One side of a regular decagon is 46. Find the area. 

C/2 = 180°/7i = 18°. 

h = (c/2) cot (C/2) = 23 cot 18°. 

p = 10 c = 460. 

F = ph/2 = 230 x 23 x cot 18°. 

log 230 = 2.36173 

log 23= 1.36173 

log cot 18° = 0.48822 

.-. log F= 4.21168 

.-. F = 16281. 

45. In a circle with a diameter of 125 ft., find the area of a sector with 
an arc of 22°. 

Area of sector = tz (62. 5) 2 22/360. 

log ?r = 0.49715 

log62.5 2 = 3.59176 

log 22 = 1.34242 

.-. log product = 5.43133 

log 360 = 2.55630 

.-. log area = 2.87503 

.-. area = 749.95. 

46. In a circle with a diameter of 50 ft. find the area of a segment with 
an arc of 280°. 

The segment is made up of a sector of 280° and an isosceles triangle 
whose vertical angle is 80° and whose equal sides are 25 ft. each. 
Area sector = 7t ■ 25 2 ■ 280/360. 
Area triangle = 25 2 sin 80°/2. 



EXERCISE XXXIII 163 

log it = 0.49715 log 25 2 = 2. 79588 

log25 2 = 2.79588 log sin 80°= 9.99335 - H > 

log 280 = 2.44716 .-. log (2 area) = 2.78923 

.-. log product = 5.74019 ... area _ 615.5/2 

log 360 = 2.55630 =307.75. 
.-. log area = 3.18389 ... area f segment = 1834.95. 
.-. area = 1527.2. 

47. A building is 37.54 ft. wide and the slope of the roof is 43° 36'. 
Find the length of the rafters. 

Here A = 43° 36', C = 90°, b = 18.77 ; to find c. 
c = 6/cos A. 
log 6= 11.27346-10 
log cos A = 9.85984 -10 
.-. logc= 1.41362 
.-. c = 25.919. 

48. What angle at the center of a circle does a chord which is 4/7 of 
the radius subtend? 

Let the radius = 7, then the chord = 4, 

and cos angle = (7 2 + 7 2 - 4 2 )/(2 -7-7) 

= .8367. 
.-. angle = 33° 12' 30". 

49. The side of a regular pentagon is 2. Find the radius of the inscribed 
circle. 

Here c/2 = 1, C/2 = 36° ; to find h. 

h = (c/2) cot (C/2) = cot (C/2). 
.-. h = 1.3764. 

50. The side of a regular decagon is 23.41 ft. Find the radius of the 
inscribed circle. 

Here c/2 = 11.705, C/2 = 18° ; to find h. 

h = (c/2) cot (C/2). 
log (c/2) = 1.06838 
log cot (C/2) = 0.48822 
.-. \ogh = 1.55660 
.-. h = 36.025. 



164 PLANE TRIGONOMETRY 

51. The perimeter of a regular polygon of 11 sides is 23.47 ft. Find 
the radius of the circumscribed circle. 

Here c = 23.47/22 = 1.0668, C/2 = 360°/22 = 16° 21' 49" ; to find r. 
r = (c/2)/sin (C/2). 
log (c/2) = 10.02808-10 
log sin (C/2) = 9.44984-10 
.-. logr = 0.57824 
.-. r = 3.7865. 

52. The perimeter of a regular heptagon inscribed in a circle is 12. 
Find the radius of the circle. 

Here c/2 = 12/14 = .85714, C/2 = 360°/14 = 25° 42' 51". 
r = (c/2)/sin (C/2). 
log (c/2) = 9.93305 -10 
log sin (C/2) = 9.63737 - 10 
.-. logr = 0.29568 
,\ r = 1.9755. 

53. Find the perimeter of a regular decagon circumscribed about a 
unit circle. 

Here h = 1, C/2 = 360°/20 = 18°; to find 10 c. 

c/2 = h tan (C/2) = tan (C/2) = .3249. 
.-. 10 c = 6.498. 

54. Find the perimeter of a polygon of 11 sides inscribed in a unit 
circle. 

Here r = 1, C/2 = 360°/22 = 16° 21' 49" ; to find 11 c. 

c/2 = r sin (C/2) = sin (C/2) = .28175. 
.-. lie = 6.198. 

55. The perimeter of an equilateral triangle is 17.2 ft. Find the area 
of the inscribed circle. 

Here (c/2) = 17.2/6 = 2.8667, C/2 = 860°/6 = 60° ; to find nW. 
h = (c/2) cot (C/2). 
.-. area = 7th 2 = tt(c/2) 2 cot 2 (C/2). 
log 7t = 0.49715 
log (c/2) 2 = 0.91476 
■ log cot 2 (C/2) = 9.52288 - 10 
.-. log area = 0.93479 
.-. area = 8.6058. 



SPHERICAL TRIGONOMETRY 



EXERCISE XXXTV 

For examples 1, 2, and 3, see the text-book. 

4. How many triangles are possible when : 

(i) A = 60°, a = 58° 31.5', 6= 80°, p = 58° 31.5'? 

(ii) A = 60°, a = 80°, b = 70°, p = 54° 28.1' ? 

(iii) A = 60=, a = 70°, b = 80°, p = 58° 31.5'? 

(iv) A = 60 : . a = 54°, b = 80°, p = 58° 31.5' ? 

In (i), a = p ; hence A ABC is right angled at B. 

In (ii), the value of a lies between p and 180° — 6, but not between p 
and 6 ; hence there is one triangle having the given parts. 

In (iii), the value of a lies between p and b and also between p and 
180° — b ; hence there are two triangles having the given parts. 

In (iv), a <p and A < 90°; hence there is no triangle having the given 
parts. 

5. By inspecting Fig. 63 determine what are the limits (1) of the sum 
of the sides of a convex spherical triangle, or of the sum of the face angles 
of a triedral angle ; (2) of the sum of the diedral angles of a triedral 
angle, or of the sum of the angles of a convex spherical triangle. 

From the figure we see that the construction of a convex spherical 
triangle is possible only when the sum of the sides is less than 360°. 

When the sum of the sides is very near 360°, each spherical angle is 
near 180°; hence the sum of the angles is less than 540°. When the sum 
of two sides is nearly equal to the third, each of two spherical angles is 
near zero and the third near to 180°; hence the sum of the angles is 
greater than 180°. 

6. Construct in cardboard a spherical polygon whose sides are 40°, 50°, 
60°. 70° respectively. Are its angles determined ? 

Like a plane quadrilateral a spherical quadrilateral is not determined 
by its sides. 

165 



166 SPHERICAL TRIGONOMETRY 

EXERCISE XXXV 

1. Write the equations for finding A, a, and c from b and B. 
To find A, we select (5), which involves A, B, 6, and obtain 

sin A = cos B/cos b. 
To find a, we select (4), which involves a, b, B, and obtain 

sin a = tan 6/ tan B. 
To find c, we select (2), which involves c, 6, 5, and obtain 

sin c = sin 6/sin B. 

2. Write the equations for finding a, A, B from 6 and c. 
From (1), cos a = cos c/cos 6. 

From (3), cos A = tan 6/tan c. 

From (2), sin B = sin 6/sin c. 

3. Write the equations for finding a, 6, 2? from A and c. 
From (2), sin a = sin A sin c. 

From (3), tan b = cos A tan c. 

From (6), cot B = tan J. cos c. 

4. Write the equations for finding A, 6, c from a and 2?. 
From (5), cos .A = cos a sin 2?. 

From (4), tan b = sin a tan 5. 

From (3), tan c = tan a/cos B. 

EXERCISE XXXVI 

1. By Napier's rules, write each of the ten equations in § 110, and 
from them write the ten equations in § 108. 
By Napier's rules, we have 

sin a = cos (co-c) cos {go- A) or tan b tan (co-Z?), (1) 

sin b — cos (co-c) cos (co-B) or tan a tan (co-A), (2) 

sin (co-A) = cos a cos (co-B) or tan b tan (co-c), (3) 

sin (co-B) = cos b cos (co-A) or tan a tan (co-c), (4) 

sin (co-c) = cos a cos b or tan (co-A) tan (co-B). (5) 

From (1), sin a = sin c sin A or tan b cot 5. (1') 

From (2), sin b = sin c sin 2? or tan a cot A. (2') 

From (3), cos A = cos a sin B or tan 6 cot c. (3') 

From (4), cos B = cos 6 sin A or tan a cot c. (4') 

From (5), cos c ='cos a cos b or cot A. cot B. (5 X ) 



EXERCISE XXXVI 167 

2. Write each of the ten equations in § 108 directly from Napier's 
rules. 

If in example 1 we do mentally the work in equations (l)-(5), we can 
write out at once equations (l / )-(5 / ). 

3. AVrite the equations for finding 6, B, and c from a and A. 

To find b from a and A, we select b as the middle part, and from Rule 
I we obtain s [ n & = tan a cot A. 

To find B from a and A, we select co-J. as the middle part, and from 
Rule II we obtain 

cos A = cos a sin J5, or sin B = cos A/cos a. 

To find c from a and A, we select a as the middle part, and we have 
sin a = sin c sin A, or sin c = sin a/sin A. 

4. Solve examples 2, 3, and 4 in Exercise XXXV. 

(2) Write the equations for finding a, A, B from b and c. 
To find a from 6 and c, we have 

cos c = cos a cos 6, or cos a = cos c/cos 6. 
To find A from 6 and c, we have 

cos A = tan 6 cot c. 
To find B from 6 and c, we have 

sin & = sin c sin 5, or sin B = sin 6/sin c. 

(3) Write the equations for finding a, 6, B from -4 and c. 
To find a from J. and c, we have 

sin a = sin A sin c. 
To find b from J. and c, we have 

cos A = tan b cot c, or tan b = cos J. tan c. 
To find 5 from A and o, we have 

cos c = cot A cot _B, cot B = tan A cos c. 

(4) Write the equations for finding J., 6, c from a and B. 
To find J. from a and B, we have 

cos A = cos a sin B. 
To find 6 from a and B, we have 

sin a = tan b cot I>, or tan b = sin a tan 5. 
To find c from a and 2?, we have 

cos B = tan a cot c, or cot c = cos B cot a. 



168 



SPHERICAL TRIGONOMETRY 



EXERCISE XXXVII 



1. Given C = 90°, b = 40°, 
c - 63° 56'; to find a, A, B. 

cos a = cos c/cos 6. 
cos vl = cot c tan 6. 
sin B = sin &/sin c. 
By §111, a < 90°, A < 90°, J3<90°. 
log cos 63° 56' = 19.64288 - 20 
log cos 40° = 9.88425 - 10 
.-. log cos a = 9.75863 - 10 
.-. a = 54° 59' 47". 
log cot 63° 56' = 9.68946 - 10 
log tan 40° = 9.92381 - 10 
.-. log cos ^4 = 9.61327 - 10 
.-. A = 65° 45' 58". 
log sin 40°= 19.80807 -20 
log sin 63° 56" = 9.95341 - 10 
.-. log sin B= 9.85466 - 10 
.-. B = 45° 41' 28". 

2. Given C = 90°, a = 134°, 
A = 105°; to find 6, B, c. 

sin b — tan a cot A. 
sin 2> = cos A /cos a. 
sine = sin a/sin ^4. 
By § 111, B < 90°, c > 90°, when 
6 < 90°. 

log tan 134° = 0.01516 
log cot 105° = 9.42805 - 10 
.-. log sin b = 9.44321 - 10 
.-. 6 1 = 16°6 / 33". 
b 2 = 163° 53' 27". 
log cos 105° = 19.41300 - 20 
log cos 134° = 9.84177 - 10 
.\ log sin B= 9.57123 .- 10 
.-. B 1 = 21° 52 7 31". 
B 2 = 158° V 29". 



log sin 134° = 19.85693 - 20 
log sin 105° = 9.98494 - 10 
.-. log sine = 9.87199 - 10 
.-. ci= 131°5r55". 
c 2 = 48° 8' 5". 

3. Show that the triangle ABC 
is impossible when G =90°, a =47°, 
A= 93°; also when C=90°, a = 52°, 
A = 90°. 

If a = 47° and A = 93°, a side 
and its opposite angle would be in 
different quadrants, which is im- 
possible by § 111; hence the triangle 
is impossible. 

By § 108, sin b — tan a cot A. 

If a = 52° and A = 90°, sin b = ; 
hence b = 0°, or 180°; if b = 0°, the 
triangle becomes an arc ; if b = 180°, 
the triangle becomes a lune. 

4. Given (7=90°, a = 70° 28', c = 
98° 18'; to find vl, B, b. 

sin A = sin a/sin c. 
cos B = tan a cot c. 
cos b = cos c/cos a. 
By §111, 4 < 90°, J5>90°, 6>90°. 
log sin 70° 28 7 = 1 9. 97426 - 20 
log sin 98° 18' = 9.99543 - 10 
.-. log sin A= 9.97883 - 10 
.-. A= 72° 15' 15". 
log tan 70° 28'= 0.45005 
log cot 98° 18' = 9.16401 - 10 
.-. log cos B = 9.61406 - 10 
.-. B= 114° 16' 50". 
log cos 98° 18' = 19.15944 - 20 
log cos 70° 28' = 9.52421 - 10 
.-. log cos b= 9.63523 - 10 
.-.6= 115° 34' 42". 



EXERCISE XXXVII 



169 



5. Given C = 90°, a = 36° 25' 30", 
b = 85° 40'; to find A, B, c. 

tan A = tan a/sin 6. 
tan i? = tan &/sin a. 
cos c = cos a cos 6. 
By § 111, A < 90°, B < 90°, c < 90°. 
log tan 36° 25' 30" = 19.86802 - 20 



log sin 85° 40' 



9.99876 - 10 



.-. log tan A = 9.86926 - 10 
.-. A = 36° 30' 12". 

log tan 85° 40' = 11. 12047 - 10 
log sin 36° 25' 30" = 9.77362 - 10 
.-. log tan B= 1.34685 

.-. B = 87° 25' 26". 

log cos 36° 25' 30" = 9.90560 - 10 
log cos 85° 40' = 8.87829 - 10 
.-. log cos c = 8.78389 - 10 
.-. c = 86° 30' 52". 



6. Given (7=90°, 
/ a = 47° 30' 40"; to 

sin J. : 

COS B : 
COS b : 

By§lll,^<90°, 

log sin 47° 30' 40" : 

log sin 120° 20' 30": 

.*'. log sin A ■ 

.'. A: 

log tan 47° 30' 40" : 
log cot 120° 20' 30" : 

.'. log COS B: 
.-. B: 

log cos 120° 20' 30": 
log cos 47° 30' 40" : 

.'. log COS 6 : 
.-. 6: 



c = 120°20'30", 
find A, B, b. 

- sin a/sin c. 
= tan a cot c. 
= cos c/cos a. 

&>90°, 5>90°. 
= 19.86771 - 20 
= 9.93603 - 10 
= 9.93168-10 
= 58° 41' 53". 
= 0.03812 
= 9.76740 - 10 
= 9.80552 - 10 

= 129° 43' 8". 
= 19.70342 -20 
= 9.82959 - 10 
= 9.87383 - 10 
= 138° 24' 30". 



7. Given C - 90°, ' c = 78° 25', 
A = 36° 42' 30"; to find a, 6, B. 

By § 111, a<90°, 6<90°, B<90°. 
sin a = sine sin A. 
tan b = cos A tan c. 
cot B = tan A cos c. 
log sin 78° 25' = 9.99106 - 10 
log sin 36° 42' 30" = 9.77652 - 10 
.-. log sin a = 9.76758 - 10 
.-. a = 35° 50' 37". 

log cos 36° 42' 30" = 9.90400 - 10 
log tan 78° 25' = 0.68832 
.-. log tan b = 0.59232 

.-. b = 75° 39' 31". 
log tan 36° 42' 30" = 9.87251 - 10 
log cos 78° 25' = 9.30275 - 10 
.-. log cot B = 9.17526- 10 
.-. B = 81° 29' 6". 

8. Given C = 90°, A = 53° 23', 
c = 108°; to find a, 6, B. 

By § 111, a<90°, 6>90°, B>90°. 
sin a = sin c sin A . 
tan b = cos A tan c. 
cot B = tan A cos c. 
log sin 108° =9.97821 - 10 
log sin 53° 23' = 9.90452 - 10 
.-. log sin a = 9.88273 - 10 
.-. a = 49° 45' 42". 

log cos 53° 23' = 9.77558 - 10 
log tan 108° = 0.48822 
.-. log tan b = 0.26380 

.-. b = 118° 34' 46". 

log tan 53° 23' =0.12894 

log cos 108° = 9.48998 - 10 

.-. log tan B = 9.61892 - 10 

.-. B = 112° 34' 45 ,/ . 



170 



SPHERICAL TRIGONOMETRY 



9. Given C = 90°, a = 122° 15', 
B=U° 20'; to find A, 6, c. 

cos 4 = cos a sin 5. 
tan b = sin a tan B. 
tan c = tan a/cos E. 

By §111,4 > 90°, 6 < 90°, c>90°. 
log cos 122° 15' = 9.72723 - 10 
log sin 14° 20' = 9.39369- 10 
.-. log cos 4 = 9.12092 - 10 
.-. A = 97° 35' 28 // . 
log sin 122° 15' = 9.92723 - 10 
log tan 14° 20' = 9.40742 - 10 
.-, log tan 6 = 9.33465 - 10 
.-. b = 12° 11' 38". 
log tan 122° 15' = 10.20000 - 10 
log cos 14° 20' = 9.98627 - 10 
.-. log tan c = 0.21373 

.-. c = 121° 26' 19". 



11. Given C = 90°,' A = 63° 18', 
B = 37° 47' ; to find a, 6, c. 

cos a = cos A/s'm B. 
cos 6 = cos B/sin A. 
cos c = cot J. cot B. 
By § 111, a < 90°, b < 90°, c < 90°. 

log cos 03° 18' = 19.65255 - 20 

log sin 37° 47' = 9.78723 - 10 

.-. log cos a = 9.86532 - 10 

.-. a = 42° 49' 50". 

log cos 37° 47' = 19.89781 - 20 
log sin 63° 18' = 9.95103 - 10 

.-. log cos b= 9.94678 - 10 
.-. b = 27° 47' 20". 
log cot 63° 18' = 9.70152 - 10 
log cot 37° 47' = 0.11058 

.-. log cose = 9.81210 - 10 

.-. c = 49° 33'. 



10. Given C = 90°, a = 108° 45', 
B = 37° 42'; to find 4, 6, c. 

cos A = cos a sin B. 
tan b = sin a tan B. 
tan c = tan a/cos B. 

By § 111, A > 90°, b < 90°, c > 90°. 
log cos 108° 45' = 9.50710 - 10 
log sin 37° 42' = 9.78642 - 10 
.-. log cos A = 9.29352 - 10 

.-. A = 101° 20' 11". 

log sin 108° 45' = 9.97632 - 10 

log tan 37° 42' = 9.88812 - 10 

.-. log tan 6 = 9.86444- 10 

.-. b = 36° 11' 58". 
log tan 108° 45' = 10.46922 - 10 
log cos 37° 42' = 9.89830 - 10 
.-. log tan c= 0.57092 
.•'. c = 105° 2' 2". 



12. Given G = 
B = 64° ; to find a. 

cos a 
cos b 
cos c 



By 



illl, a>90 c 

log cos 106° 

log sin 64° 

.-. log cos a 

.-. a 

log cos 64° 

log sin 106° 

.-. log cos b 

.-. b 

log cot 106° 

log cot 64° 

.-. log cose 

.-. c 



90°, A = 106°, 
6, c. 

= cos A/sin B. 
= cos B/sin A. 
= cot A cot B. 
, b < 90°, c > 90°. 
= 19.44034 - 20 
= 9.95366 - 10 



9.48668 - 


- 10 


107° 51' 32". 


19.64184 - 


- 20 


9.98284 - 


- 10 


9.65900 - 


- 10 


62° 52' 5" 




9.45750 - 


10 


9.68818 - 


10 



14568 - 10 
S° 2' 22". 



EXERCISE XXXVII 



171 



13. Given C = 90°, a = 47° 40', 
A = 30° 43' ; to find 6, B, c. 

sin b = tan a cot A. 
sin c = sin a/sin ^4. 
sin B = cos ^4 /cos a. 

tana>l, cot.4>l; .-. sin6>l. 
sin a > sin A; .-. sinc>l. 

cos^4>cosa; .-. sinl?>l. 

Hence the triangle is impossible. 

14. Given C = 90°, a = 64° 29', 
A = 78° 10' ; to find 6, 5, c. 

sin& = tan a cot .4. 
sin B = cos ^4/cos a. 
sin c = sin a/sin J.. 

By §111, .B<90 o , c<90°, when 
b < 90°. 

log tan 64° 29' =0.32118 
log cot 78° 10' = 9.32122 - 10 
.-. log sin b = 9.64240 - 10 

.-. &i = 26° 2' 9". 
6 2 = 153° 57' 51". 

log cos 78° 10' = 19.31189 - 20 

log cos 64° 29' = 9.63425 - 10 

.-. log sin B= 9.67764 - 10 

.-. J?i = 28° 25' 37". 
B 2 = 151° 34' 23". 

log sin 64° 29'= 19.95543 - 20 

log sin 78° 10' = 9.99067 - 10 

.-. log sin c= 9.96476- 10 

.-. ci = 67° 13' 48". 
c 2 = 112° 46' 12". 

15. Given C = 90°, B = 47° 50', 
c = 78°20'; to find A, a, b. 



tan A - 
tan a = 
sin 6 = 

By § 111, A< 90°, 

log cot 47° 50' = 

log cos 78° 20' = 

.-. log tan A - 

.-. A-- 

log cos 47° 50' = 

log tan 78° 20' = 
.-. log tan a - 



log sin 78° 20' 
log sin 47° 50' 

.-. log sin b 



= cot J5/cos c. 
= cos B tan c. 
= sin c sin B. 

a < 90°, b < 90° 

= 9.95698 - 10 
= 9.30582 - 10 
= 0.65116 

i 77° 24' 50". 

i 9.82691 - 10 
: 0.08511 
: 0.51202 

: 72° 54' 8". 

: 9.99093 - 10 
: 9.86993 - 10 
: 9.86086 - 10 

: 46° 32' 30". 



16. Given C = 90°, c = 84° 47', 
b = 39° 43'; to find a, A, B. 

cos a = cos c/cos b. 
cos A = tan b cot c. 
sin B = sin 6/sin c. 

By § 111, a < 90°, A < 90°, J5<90°. 

log cos 84° 47' = 18.95867 - 20 

log cos 39° 43' = 9.88605 - 10 

.-. log cos a = 9.07262 - 10 

.-. a = 83° 12' 42". 

log tan 39° 43' = 9.91945 - 10 

log cot 84° 47' = 8.96047 - 10 

.-. log cos A = 8.87992 - 

.-. A = 85° 39' V 



10 



log sin 39° 43' = 19.80550 - 20 

log sin 84° 47' = 9.99820 - 10 

.-. log sin B= 9.80730 - 10 

.-. B= 39° 54' 56". 



172 



SPHERICAL TRIGONOMETRY 



17. Given C = 90°, A = 124° 30', 
b = 25° 40' ; to find a, B, c. 

tan a = tan A sin b. 
cos B— sin 4 • cos 6. 
tan c == tan 6/cos J.. 

By §111, a > 90°, £<90°, c>90°. 

log tan 124° 30' = 0.16287 
log sin 25° 40' = 9.63662 - 10 
.-. log tan a = 9.79949 - 10 

.-. a = 147° 46' 49". 

log sin 124° 30' = 9.91599 - 10 

log cos 25° 40' = 9.95488 - 10 

.-. log cos B =9.87087 - 10 

.-. B = 42° r 49". 

log tan 25° 40' = 19.68174 - 20 

log cos 124° 30' = 9.75313 - 10 

.-. log tan c= 9.92861 - 10 

.-. c = 139° 4r 17". 

18. Given C = 90°, b = 106°, 
B = 100° ; to find a, A, c. 

sin a = tan b cot B. 

sin A = cos .B/cos &• 

sin c = sin &/sin B. 

By § 111, A < 90°, c > 90°, when 
a < 90°. 

log tan 106° = 0.54250 

log cot 100° = 9.24632 - 10 

.-. log sin a = 9.78882 - 10 

ai = 37° 56' 46". 
a 2 = 142° 3 X 14". 

log cos 100° = 19.23967 - 20 
log cos 106° = 9.44034 - 10 
.-. log sin A= 9.79933 - 10 

.-. ^ 1 = 39°2 / 56". 
A 2 = 140° 57' 4". 



log sin 106° = 19.98284 - 20 
log sin 100° = 9.99335 - 10 
.-. log sin c= 9.98949- 10 

ci = 103° 33' 20". 

c 2 = 76° 26 7 40". 

19. Given C = 90°, A = 76° 15', 
B = 49° 3' ; to find a, 6, c. 

cos a = cos A/sin B. 
cos b = cos B/s'm A. 
cos c = cot A cot B. 
By § 111, a < 90°, b < 90°, c < 90°. 
log cos 76° 15' = 19.37600 - 20 
log sin 49° 3' = 9.87811 - 10 
.-. log cos a = 9.49789 - 10 
.-. a = 71° 39' 27". 
log cos 49° 3' = 19.81651 - 20 
log sin 76° 15' = 9.98737 - 10 
.-. log cos b= 9.82914 - 10 
.-. b = 47° 33' 56". 
log cot 76° 15'= 9.38863- 10 
log cot 49° 3' = 9.93840 - 10 
•. log cos c = 9.32703 - 10 
.-. c = 77° 44' 26". 

20. Write all the different groups 
of three elements that can be formed 
from the five elements a, 6, c, A, B. 

abc, abA, abB, acA, acB, aAB, 
be A, bcB, bAB, cAB. 

21. Write the formula which in- 
volves the elements of each group in 
example 20. 

The ten formulas, each of which 
involves one of the ten groups in 
example 20, are the ten formulas 
in § 108 for solving right-angled 
triangles. 

Hence we have a formula for 
every group of three of the five 
elements a, 6, c, A, B. 



EXERCISE XXXVIII 



173 



EXERCISE XXXVIII 



1. Given a = 66° 32', b = 59° 43', 
c = 90°; to find A, B, C. 

In the polar triangle A'B'C we 
have A' = 113° 28', B' = 120° 17', 
C = 90° ; to find a', 6', c'. 

cos a' = cos J. '/sin B'. 

cos 6' = cos .BVsin A'. 

cos c' = cot B' cot J/. 

By§lll,a'>90°, &'>90°, c / <90°. 

log cos A' = 19.60012 - 20 

log sin B' = 9.93628 - 10 

log cos a' = 9.66384 -10 

.-. a' = 117° 27' 40". 
log cos 5' =19.70267 -20 
log sin A' = 9.96251 - 10 
.-. log cos 6' = 9.74016 -10 
..-. b' = 123° 20' 57". 
log cot 4' = 9.63761 -10 
log cot B' = 9.76639-1 
.-. logcosc' = 9.40400 -10 
.-. c' = 75° 18' 53". 
... a = 180° - a' = 62° 22' 20". 
B = 180° - 6' = 50° 39' 3". 
C = 180° - c' = 104° 41' 7". 

2. Given a = 123° 48' 24", 
C = 67° 12', c = 90°; to find A, B, b. 

In the polar triangle A'B'C we 
have .4'= 56° 11' 36", c'=112° 48', 
C" = 90° ; to find a', &', J3'. 

sin a' = sin c' sin .4'. 

tan b' = cos .4/ tan c'. 

cot J3' = cos c' tan .4'. 

By § 111, a'<90°, B'> 90°, &'> 90°. 

logsinc / = 9.96467-10 

log sin A' = 9.91956-10 

.-. logr sin a' = 9.88423- 10 



.-. a' 

log cos c' 

log tan A' 

. log cot B' 

.-. B' 

log cos ^4' 

log tan c' 

\ log tan b' 

4 = 180°- 
£ =r 180° - 
6 = 180° - 



= 49° 59' 48". 
= 9.58829-10 
= 0.17418 



9.76247- 


10 


120° 3' 32 




9.74538- 


10 


0.37638 





= 0.12176 
= 127° 4' 16". 
a'=130°0' 12" 
b' = 52° 55' 44" 
B' = 59° 56' 28' 



3. Given C = 136° 4', 5 = 140°, 
c = 90°; to find a, &, A. 

In the polar triangle A'B'C we 
have c'= 43° 56', b' = 40°, C" = 90° ; 
to find A', B\ a'. 

cos A' = tan b' cot c'. 
cos a' = cos c'/cos b'. 
sin .B' = sin &'/sin c'. 
By § 111, 3'<90°, a'<90°, A'<90°. 
log tan 6'= 9.92381 -10 
log cot c' = 0.01617 
.-. log cos 4' = 9.93998 -10 
.-. 4' =29° 26'. 
log sin b' = 19.80807 -20 
log sin c' = 9.84125 -10 
.-. log sin B' = 9.96682 - 10 
.-. 5' =67° 53' 12". 
log cose' = 19.85742 -20 
log cos b' = 9.88425 - 10 
.-. log cos a' = 9.97317 - 10 
.-. a' = 19° 56'. 
.-. a = 180° -4' =150° 34'. 
b = 180° - B' = 112° 6' 48". 
A = 180° - a' = 160° 4'. 



174 SPHERICAL TRIGONOMETRY 

4. Given b = c = 81° 24', A = 72° 40'; to find a, £, C. 

From A draw AM ± CB; then in the right spherical triangle CAM we 
have (L4 = b = 81° 24', Z C^LJf = .4/2 = 36° 20', Z CJfA = 90°; to find 
C and CM, or a/2. cot o = cos 6 tan (A/2). 

sin (a/2) = sin b sin (-4/2). 
log cos 6 = 9. 17474 - 10 log sin 6 = 9. 99509 - 10 

log tan (A/2) = 9.86656-10 log sin (A/2) = 9.77268-10 

.-. log cot C = 9.04130 - 10 .-. log sin (a/2) = 9.76777 - 10 

.-. C = 83° 43' 26" = B. .: a/2 = 35° 51' 42"+. 

.-. a = 71° 43' 25". 

5. Given a = b = 54° 20', c = 72° 54'; to find A, B, C. 

Draw CM±AB; then in the right spherical triangle .CMB we have 
CB = a = 54° 20', MB = c/2 = 36° 27'; to find B and C/2. 
cos J5 = cot a tan (c/2). 
sin (C/2) = sin (c/2)/sin a. 
log cot a = 9. 85594 - 10 log sin (c/2) = 19. 77387 - 20 

log tan (c/2) = 9.86842-10 log sin a = 9.90978 -10 

.-. log cos B = 9.72436-10 . .\ log sin (C/2) = 9.86409-10 

.-. B = 57° 59' 15" = A. .-. C/2 = 46° 59' 40". . 

.-. C = 93° 59' 20". 

6. Given a = b = 54° 30', C = 71° ; to find ^, B, c. 

Draw Clf±7lJ5; then in the right spherical triangle CMB we have 
C/2 = 35° 30', CB=a = 54° 30' ; to find B and c. 
cot -B = cos a tan (C/2). 
sin (c/2) = sin a sin (C/2). 
log cos a = 9. 76395 - 10 log sin a = 9. 91069 - 10 

log tan (C/2) = 9.85327 - 10 log sin (C/2) = 9.76395 - 10 

.-. log cot B = 9.61722 - 10 .-. log sin (c/2) = 9.67464 - 10 

.-. B = 67° 30' = A. .-. c/2 = 28° 12 r 50". 

.-. c = 56° 25 y 40". 

7. Given a = b, c = 116° 40', C = 127° 46'; to find A, B, a, b. 

Draw CM±AB; then in the right spherical triangle CMB we have 
C/2 = 63° 53', MB = c/2 = 58° 20', Z CMB = 90° ; to find B and CB, or a. 
sin a = sin (c/2)/sin (C/2). 
sin B = cos(C/2)/cos(c/2). 
log sin 58° 20' = 19.92999 - 20 
log sin 63° 53' = 9.95323 - 10 
.-. log sin a = 9.97676 -10 



EXERCISE XXXIX 



175 



.-. a x = 71° 25' 24", a 2 = 108° 34' 36' 
log cos 63° 53' = 19.64365 - 20 
log cos 58° 20' = 9.72014 - 10 
log sin B = 9.92351 -10 

.-. B x = 56° 59', B 2 = 123° V. 



EXERCISE XXXIX 

1. Write the equations for finding the three angles from the three sides. 
By [42], cos A = (cos a — cos b cos c)/(sin b sin c). 

By [41], sin B = sin b sin A /sin a, sin C = sin c sin A /sin a. 

2. Write the equations for finding the three sides from the three angles. 
By [43], cos a = (cos A + cos B cos C)/(sin B sin C). 

By [41], sin b = sin a sin D/sin -4., sin c = sin a sin C/sin J.. 

3. Write the equations for finding A, B, c from a, 6, C. 
By [42], cos c = cos a cos b + sin a sin b cos C. 

By [41], sin A — sin a sin C/sin c, sin B = sin 6 sin C/sin c. 

4. Write the equations for finding A, b, c from B, C, a. 
By [43] , cos A = — cos B cos C + sin B sin C cos a. 

By [41], sin 6 = sin a sin J3/sin A, sin c = sin a sin C/sin -4. 

5. Write the equations for finding A, C, c from a, 6, 5. 
By [41], sin ^4 = sin a sin J5/sin b. 

By [42], cos b = cos c cos a + sin c sin a cos B. (1) 

To solve (1 ) for cos c or sin c would involve radicals ; hence we deduce 
later simpler formulas for this case. 

By [41], sin C = sin B sin c/sin b. 

6. Prove the law of sines geometrically when each side is less than 90°. 
Take OP = 1. Draw PD ± OCB, DE ± 

OB, and DF±OC, and join PE and PF; 
then P^ JL OB, and PP J_ OC. § 104 

.-. Z DPP = C, 

and Z DPP = B ; 

hence in A DFP, sin C = DP/sin 6, 

and in A DEP, sin P = DP/sin c. 

.-. sin P/sin C = sin 6/sin c, 

or sin 6/sin B = sin c/sin C. Fig. 14 




176 



SPHERICAL TRIGONOMETRY 



7. Prove the law of cosines geometrically when each side is less than 90°. 
On OA take OP = 1. In C OA draw PD JL OA at P. In BOA draw 

PE _L (M at P ; then Z DPP = J., and Z COP = a. 

In A DOE, 
DE 2 = sec 2 6-h sec 2 c — 2 sec b sec c cos a. (1) 

In A DPE, 
DP 2 = tan 2 &+tan 2 c-2tan&tanccos^4. (2) 

Subtracting (2) from (1) and using [5], 
we obtain q 

= 2 — 2 sec b sec c cos a 
+ 2 tan b tan c cos J.. 
.-. cos a = cos b cos c + sin b sin c cos ^4. 

If on OB we take OP = 1, and repeat the construction and proof as 
above, we obtain the value of cos b. By taking OP = 1 on OC, we obtain 
cose. 

8. Draw the figures and deduce the formulas for cos b and cos c. 

In fig. x, B and C are both acute or both obtuse. In fig. y, B is obtuse 
and C is acute. 






D *■*—-— _ 



Fig. 16 

Regarding 5D and DC as directed arcs, we have 

a = BD + DC, or DC = a - BD. 
108, tan BD = tan c cos 5, 

cos 6 = cos DC cosp 

= cos (a — BD)cosp 

= cos a cos PD cosp + sin a sin PD cosp. (3) 
By § 108, cos BD cosp = cos c. (4) 

Whence cosp = cos c/cos BD. 

.-. sin PD cosp = cose tan BD 

= cos c tan c cos P = sin c cos P. (5) 



By 

and 



(1) 
(2) 



EXERCISE XXXIX 177 

From (3), by (4) and (5), we obtain 

cos b = cos a cos c + sin a sin c cos B. 
Interchanging B and C and also b and c in the figures, and proceeding 
as above, we obtain the value of cos c. 

9. Derive the formulas for cos B and cos C from those for cos b and 
cos c by means of polar triangles. 

Let A'B'C be the polar triangle of ABC. 

By [42], cos b' = cos a' cos of + sin a' sin c' cos B 7 , 

and cos c' = cos a' cos 6' + sin a' sin 6' cos C". 

Substituting for a', &', c', B\ and C their equals, 180° - A, 180° - B, 
180° - C, 180° - 6, 180° - c respectively, we obtain 

cos (180° - B) = cos (180° - A) cos (180° - C) 

+ sin (180° - A) sin (180° - C) cos (180° - &), 
and cos (180° - C) = cos (180° - A) cos (180° - B) 

+ sin (180° - J.) sin (180° - C) cos (180° - c). 
.-. cos B = — cos J. cos C + sin A sin C cos 6, 
and cos C = — cos J. cos 5 + sin A sin 2? cos c. 

10. Deduce (1), (2), (5), (6) in § 108 by putting C = 90° in the law of 
sines and the law of cosines. 

Putting C = 90° in [41], we obtain 
sin A = sin a/sin c, 1 

sin 5 = sin 6/sin c. J ' - 

Putting C = 90° in the value of cos c, by [42], we obtain 

cose = cos a cos b. (1) 

Putting C - 90° in [43], we obtain 

cos A = sin B cos a, 1 

cos B = sinA cos 6, j ^ ' 

and = — cos A cos B + sinA sin 5 cos c, 

or cos c = cot A cot J5. (6) 

11. Prom [42] and (1), in § 108, deduce (3) in § 108. 
Prom [42] and (1), in § 108, we obtain 

cos c cos b cos c 



cos A — 



sin b cos b sin c sin b sin c 
cos6\ 



cote 



\sin b < 



cos 6 sin b 



1 _ cos 2 6 
cot c = cot c tan 6, or (3). 

sin b cos 6 



178 SPHERICAL TRIGONOMETRY 

12. Deduce (4) in § 108 by dividing (2) by (3) and then using (1). 

Dividing (2) by (3), we obtain 

sin a tan c 

tan A = 

sin c tan b 

_ sin a sin c cos b 

sine cose sin 6 

sin a cos b 



by(l) 



cos a cos b sin b 
= tan a/sin b. 

EXERCISE XL 

1. Write the formulas for finding A, B, C from a, b, c. 

tan (A/2) = tan r/sin (s — a). 
tan (-B/2) = tan r/sin (s — 6). 
tan (C/2) = tan r/sin (s — c). 
A, B,C can also be found through the sines or cosines of A/2, B/2, C/2. 

2. Write the formulas for finding a, 6, c from A, B, C. 

tan (a/2) = tan E cos (S — -4). 
tan (6/2) = tan R cos (S - B). 
tan (c/2) = tan R cos (5 - C). 

3. Write the formulas for finding A, B, c from a, b, O. 

tan i (J. + 5) = cos i (a - 6) cot (C/2)/eos \ (a + b). 
tan i (J. - B) = sin | (a — b) cot (C/2)/sin \ (a + b). 
sin c = sin a sin C/sin J.. 

4. Write the formulas for finding a, b, C from A, B, c. 

tan i (a + 6) = cos \ (A - B) tan (c/2)/cos J (^1 + £). 
tan i (a - 6) = sin f (.4 - 5) tan (c/2)/sin \(A + B). 
sin C = sin J. sin c/sin a. 

5. Write the formulas for finding J?, C, c from a, 6, ^4. 

sin B = sin 6 sin .A /sin a. 
cot (C/2) = sin i (a + 6) tan | (A - 5) /sin \ (a - b). 
tan (c/2) = sin \ (A + B) tan | (a'- 6)/sin ±(A - B). 

6. Write the formulas for finding &, c, C from A, B, a. 

sin 6 = sin 5 sin a/sin A. 
cot (0/2) = sin i (6 + a) tan | (5 - J.)/sin | (b - a). 
tan (c/2) = sin \ (B + A) tan i (6 - a)/sin \ (B - J.). 



EXERCISE XLI 179 

7. State Napier's analogies in words. 

The sine of half the sum of two angles is to the sine of half their differ- 
ence as the tangent of half their included side is to the tangent of half the 
difference of their opposite sides. 

The cosine of half the sum of two angles is to the cosine of half their 
difference as the tangent of half their included side is to the tangent of 
half the sum of their opposite sides. 

The sine of half the sum of two sides is to the sine of half their differ- 
ence as the cotangent of half their included angle is to the tangent of half 
the difference of their opposite angles. 

The cosine of half the sum of two sides is to the cosine of half their 
difference as the cotangent of half their included angle is to the tangent 
of half the sum of their opposite angles. 

8. Write the analogies involving tan (6/2); cot (A/2). 

sin \ (A + C) _ tan (6/2) cos |( A + C) ■ . tan (6/2) 



sin | (yi — C) tan \(a — c) cos i (A — C) tan \ (a + c) 
sini(6 + c) cot (.4/2) cos|(6+c) cot (A/2) 



sin i (b — c) tan \ (B - C) cos h(b - c) tan \(B + C) 

9. When a + 6 > 180° and B < 90°, then A > 90°. 

In [49] , tan (c/2) and cos \(A — B) are both positive ; hence cos \ ( A + B) 
and tan \ (a + 6) are both positive or both negative ; that is, when 
a + 6 > 180°, B + A > 180° ; hence if a + 6 > 180° and B < 90°, A > 90°. 

10. When A + C < 180° and a > 90°, then c < 90°. 

When A + C < 180°, a + c < 180° : hence, if a > 90°, c < 90°. 



EXERCISE XLI 

1. Given a = 58° 18', 6 = 62° 40', c = 78° 24' ; to find A, B, C. 



tan r = Vsin (s — a) sin (s — 6) sin (s — c)/sin 
tan (A/2) = tan r/sin (s — a). 
tan (B/2) = tan r/sin (s — 6). 
tan (C/2) = tan r/sin (s — c). 

Here 2 s = 199° 22', .-. s = 99° 41', 

s - a = 41° 23', s - 6 = 37° 1', s - c = 21° 17'. 



180 



SPHERICAL TRIGONOMETRY 



log sin (s-a) = 9.82026 - 10 

log sin (s - 6) = 9. 77963 - 10 

log sin (s - c)= 9.55988 -10 

.-. log product = 19.15977 - 20 

log sin s= 9.99377 -10 

.-. log tan r = (19. 16600 - 20)/2 

= 19.58300 -20. 



log tan (A/2) = 9.76274 - 10. 
log tan (B/2) = 9.80337 - 10. 
log tan (C/2) = 0.02312. 
.-. A/2 = 30° 4' 27 // . 

B/2 = 32° 27' 4". 

0/2 = 46° 31' 28". 



2. Given a = 110° 4', b = 74° 32', c = 56° 30'; to find A, B, C. 

tan r = Vsin (s 



s = 120° 33'. 




s - a = 10° 29'. 




log sin (s - a) = 9.25995- 


-10 


logsin(s-6)= 9.85706- 


-10 


log sin (s-c)= 9.95384- 


-10 


•. log product = 19.07085- 


-20 


logsins= 9.93510- 


-10 



.-. log tan r = (19. 13575 - 20)/2 
= 19.56788-20. 



a) sin (s — b) • sin (s — c)/sin 
tan (-4/2) = tan r/sin (s — a). 
tan (B/2) = tan r/sin (s — b). 
tan (C/2) = tan r/sin (s — c). 

s - 6 = 46° Y. 
s-c = 6i° 3'. 
log tan (A/2) = 0.30793. 
log tan (B/2) = 9.71082 - 10. 
log tan (C/2) = 9.61404 - 10. 
.-. ^4/2 = 63° 47' 51". 
J5/2 = 27°11'45". 
C/2 = 22° 21' 7". 



3. Given A = 53° 50', B = 96° 18', C = 64° 20' ; to find a, 6, c. 

— cos S 



tan E 



V; 



cos (S - ^1) ■ cos (S - B) cos (5 - C) 
tan (a/2) = tan B cos (£ — A), etc. 

5 = 107° 14'. S - B = 10° 56'. 

8 - ^4 = 53° 24'. S - C = 42° 54'. 



9.77541 -10 
9.99204 - 10 
9.86483 - 10 



log cos (S — A) — 
log cos (S - B) = 
log cos (S-C) = 
.-. log product = 
log cos S = 
.-. log tan R - (19783940 



9.63228 
19.47168 



10 
20 



20)/2 

91970 - 10. 



log tan (a/2) = 9.69511 - 10. 
log tan (b/2) = 9.91174 - 10. 
log tan (c/2) = 9.78453 - 10. 
.-. a/2 = 26° 21' 43". 
6/2 = 39° 13' 5". 
c/2 = 31° 20' 11". 



4. Given A = 74° 40', B = 67° 30', C = 49° 50' ; to find a, 6, c. 
tan E = V- cos S/[cos (S - A) cos (S - B) cos (6' - C)]. 
tan (a/ 2) = tan i2 cos (S — A), etc. 



EXERCISE XLI 



181 



S = 96°. 
S -A = 21° 20'. 
log cos (S - A) = 9.96917 
log cos (S - B) = 
log cos (S - O) = 
.-. log product = 
log cos £ = 
.-. log tan R = 



9.94390 
9.84046 



10 

10 
10 



9.75353 -10 
19.01923 - 20 



= (19.26570 
= 9.63285 



20)/2 
10. 



S-B = 
S-G = 

log tan (a/2) = 
log tan (6/2) = 
log tan (c/ 2) = 

.-. a/2 = 
6/2 = 
c/2 = 



28° 30'. 
46° 10'. 
9.60202 - 10. 
9.57675 - 10. 
9.47331 - 10. 
21° 47' 58". 
20° 40' 27 // . 
16° 33' 41". 



Given a = 43° 30', 6 = 72° 24', c = 87° 50' ; to find C. 
sin (C/2) = Vsin (s — a) sin (s — 6)/ (sin a sin 6). 



s = 101° 52'. 
s - a = 58° 22'. 
log sin (s - a) = 9.93014 - 10 
log sin (s - h) = 9.69189-10 
log numerator 



s - b = 29° 28'. 
s - c = 14° 2 r . 
log sin a = 9.83781 -10 
log sin 6 = 9.97918 - 10 
log denominator = 9.81699 — 10 



19.62203 - 20 
.-. log sin (C/2) = (19.80504 - 20)/2 
= 9.90252-10. 
.-. C = (53° V 48") • 2 = 106° 3' 36". 

6. Given a = 110° 40 7 , 6 = 45° 10', c = 73° 30' ; to find A. 

cos {A/2) = Vsin s sin (s — a)/(sin 6 sin c). 
s = 114°40 / . 
log sin s= 9.95844-10 
log sin (s - a) = 8.84358-10 
. log numerator = 18.80202-20 
.-. log cos (A/2) = (18.96954 - 20)/2 
= 9.48477 - 10. 
.-. A = (72° 13' 19") • 2 = 144° 26' 38". 

7. Given A = 80° 40', B = 116° 20', C = 92° 20' ; to find 

tan (6/2) = V- cos £ • cos (S 
S = 144° 40'. 
S - A = 64°. 
log cos 5 = 9.91158 -10 
log cos (S - B) = 9.94458 - 10 



s - a = 4°. 
log sin 6 = 9.85074 -10 
log sin c = 9.98174 - 10 
log denominator = 9.83248 - 10 



log numerator = 9.85616 - 10 



5)/ [cos (S-A) cos 
S-B = 
S-C = 
log cos (S — A) = 
log cos (S - C) = 
. log denominator = 



(S-C)]. 
28° 20'. 
52° 20'. 
9.64184 - 10 
9.78609 - 10 
9.42793 - 10 



log tan (6/2) 

.-. 6 = (58 



0.42823/2 
0.21412. 



55' 3") • 2 = 117° 10' 



182 



SPHERICAL TRIGONOMETRY 



EXERCISE XLII 

1. Given a = 64° 24', b = 42° 30', C = 58° 40' ; to find A, B, c. 

tan \ (A + B) = cos \ (a - b) : cot (0/2) /cos \ (a + b). 
tan ! (J. - -B) = sin i (a - b) cot (C/2)/sin | (a + 6). 
sin c = sin a • sin C/sin J. . 
(a + 6)/2 = 53° 27'. 
(a - 6)/2 = 10° 57'. 
C/2 = 29° 20'. 



log cos \ (a - b) = 9.99202-10 

log cot (C/2) = 0.25031 

.-. log product = 10.24233 - 10 

log cos § (a + b) = 9.77490-10 

log tan i (A + B)= 0.46743 

.-. (A + B)/2 = 71° 10' 41". 



log sin i (a - b) = 9.27864-10 

log cot (C/2) = 0.25031 

.-. log product = 19.52895 - 20 

log sin \ (a + b) = 9.90490-10 

log tan i (^4 -5)= 9.62405-10 

.-. {A - B)/2 = 22° 49' 12". 



.-. A = 93° 59' 53", B = 48° 21' 29". 

log sin a = 9.95513 -10 

log sin C = 9.93154 -10 

.-. log product = 9.88667 - 10 

' log sin A = 9 .99894-10 

.-. log sin c = 9.88773 -10 

.;. c = 50° 33' 6". 

By condition II, in § 120, c < a, since, C < A. 

2. Given c = 78° 15', b = 56° 20', A = 120°; to find C, S, a. 

tan i (C + B) = cos § (c - 6) cot (.4/2)/cos I (c + 6). 
tani(C-5) = sin|(c- 6)cot(^4/2)/sini(c + 6). 
sin a = sin c sin A/sin C. 
(c + 6)/2 = 67° 17' SO' 7 , 
(c - b)/2 = 10° 57' 30". 
A/2 = 60°. 



log cos I (c - 6).= 9.99201 - 10 

log cot (A/2) = 9.76144 -10 

.-. log product = 9.75345 - 10 

log cos i (c + b) = 9.58663-10 

logtani(C+J5) 

.-. (C + B)/2 



0.16682 
55° 44' 3( 



log sin i (c - 6) = 9.27897-10 

log cot (A/2) = 9.76144 -10 

.-. log product = 19.04041 - 20 

log sin \ (c + b) = 9.96496 -10 

.-. log tan \(C -B) = 9.07545 - 10 

.-. (C-£)/2 = 6°47'5". 



C = 62° 31' 41", B = 48° 57' 31". 



EXERCISE XLII 183 

log sin c= 0.99080-10 

log sin A = 9.93753 - 10 

.-. log product = 19.92833 - 20 

log sin C = 9.94804-10 

.-. log sin a = 9.98029- 10 

.-. a = 107° 8'. 

By condition II, in § 120, a > c, since A > C. 

3. Given b = 52° 12' 5", c = 54° 34', A = 97° 56' 28" ; to find B, C, a. 

tan i (O - J?) = sin | (c - 6) cot (^/2)/sin | (c + 6). 
tan i (C + J5) = cos |(c - 6) cot (vi/2)/cos |(c + 6). 
sin (a/2) = V— cos <S cos (S — ^l)/(sin jBsinC). 
i(c + 6) = 53° 23' 3". 
i (c - 6) = 1° 10' 57". 
^1/2 = 48° 58' 14". 
log sin J (c - 6) = 8.31464 - 10 log cos | (c - 6) = 9.99991 - 10 

log cot (A/2) = 9.93961 - 10 log cot (^1/2) = 9.93961 - 10 

.-. log product = 8.25425 - 10 .-. log product = 9.93952 - 10 

log sin * (c + 6) = 9.90452 - 10 log cos ^(c + 5) = 9.77557 - 10 

logtani(C-5) = 8.34973- 10 .-. log tan i(C + B) = 0.16395- 
.-. (C - B)/2 = 1° 16' 54". .-. (C + B)/2 = 55° 34'. 

.-. C = 56° 50' 54", B = 54° 17' 6". 
£ = 104° 32' 14". 
S - ^L = 6° 35' 46". 
log cos S= 9.39969-10 log sin B = 9.90952 - 10 

log cos (S -A)= 9.99711 - 10 log sin C = 9.92284-10 

log numerator = 19.39680 - 20 log denominator = 9.83236 - 10 

.-. log sin (a/2) = (19.56444 - 20)/2 = 9.78222 - 10. 
... a = (37° 16' 31") • 2 = 74° 33' 2". 
By the law of sines we obtain a — 74° 33'. 
We use the formula above for a, to determine in what quadrant a is. 

4. Given B = 98° 30', C = 67° 20', a = 60° 40' ; to find &, c, A. 

tan i (6 + c) = cos |(B - C) tan (a/2)/cos | (B + C). 
tan i (& - c) = sin | (5 - C) tan (a/2)/sin | (B + C). 
sin ^4 = sin B sin a/sin 6. 
(5 + C)/2 = 82° 55'. 
(B - C)/2 = 15° 35'. 
a/2 = 30° 20'. 



184 



SPHERICAL TRIGONOMETRY 



log cos \ (B - 0) = 9.98373 - 10 

log tan (a/ 2) = 9.76725-10 

.-. log product = 9.75098 - 10 

log cos I (B + C) = 9.09101-10 

. log tan £ (6 + c) = 0.65997 

.-. (6 + c)/2 = 77° 39' 31". 
.-. 6 = 86° 39' 34", c 



log sin £ (5 - C) = 9.42917 - 10 

log tan (a/2) = 9.76725 -10 

\ log product = 19. 19642 - 20 

log sin | (B+C)= 9.99667 - 10 

.-. log tan i (6 - c) = 9.19975 - 10 

«■. (b - c)/2 = 9° 0' 3". 
68° 39' 28". 



^ 



log sin B= 9.99520-10 
log sin a = 9.94041 - 10 

log product = 19.93561 - 20 
log sin b = 9.99926 -10 

.-. log sin A= 9.93635 -10 

.-. A = 59° 43' 53". 



By condition II, in § 120, A < C, since a < c. 



5. Given A = 125° 20', C = 48° 30', b = 83° 13' ; to find a, c, 5. 

sin i (A - C) tan (6/2) /sin i (4 + C). 
cos i (^. - C) tan (6/2)/cos i (.4 + C). 



tan i (a — c) : 

tan i (a + c) = 

sin (5/2) : 

(^ + C)/2 = 

(A - C)/2 = 

6/2 = 

log sin J(A - C) = 9.79335 - 10 

log tan (6/2) = 9.94847-10 

.-. log product = 9.74182 - 10 

log sin i (A+ C) = 9.99937 - 10 

log tan i (o - c) = 9. 74245 - 10 

.-. (a - c)/2 = 28° 55' 38". 
.-. a = 114° 30' 26", 
s - a = 12° 40' 52". 
s - c = 70° 32' 8". 



= V sin (s — a) sin (s — c)/(sin a sin c). 
z 86° 55'. 
= 38° 25'. 
r 41° 36' 30". 

log cos l(A - C) = 9.89405 - 10 

log tan (6/2) = 9.94847 - 10 

.-. log product = 9.84252 - 10 

log cos i ( A + ■€)'= 8.73069-10 

.-. log tan i (a + c) = 1.11183 

.-. (a + c)/2 = 85° 34' 48". 
c = 56° 39' 10". 



log sin (s — a) = 9.34148 — 10 log sin a 

log sin (s — c) = 9.97445 — 10 log sin c 

•. log numerator == 19.31593 — 20 .-. log denominator 
.-. log sin (5/2) = (19.43506 - 20)/2 = 9.71753 - 10. 
.-. B = (31° 27' 18") • 2 = 62° 54' 36". 

By the law of sines we obtain B = 62° 54' 34". 



9.95900-10 
9.92187-10 
9.88087-10 



EXERCISE XLIII 



185 



6. Given A = 67° 30', B = 45° 50', c = 74° 20' ; to find a, 6, C. 

tan i (a + b) = cos \{A-B) tan (c/2)/cos i(J. + B). 
tan i (a - 6) = si n * (A - B) tan (c/2)/sin \(A + B ). 
sin (C/2) = Vsin (s — a) sin (s — 6)/(sin a sin 6). 
(A + B)/2 = 56° 40'. 
{A - B)/2 = 10° 50'. 
c/2 = 37° 10'. 



log cos i (A - B) = 9.99219 - 


-10 


logsm%{A-B) = 9.27405- 


-10 


log tan (c/2) = 9.87974 - 


-10 


logtan(c/2)= 9.87974- 


-10 


.-. log products 9.87193- 


-10 


.-. log product = 19.15379 - 


-20 


log cos %(A + B) = 9.73997 - 


-10 


log sin i ( A + B) = 9.92194 - 


-10 


log tan i (a + 6) = 0.13196 




.-. log tan i (a - 6) = 9.23185 - 


-10 



(a + 6)/2 = 53° 34' 25". 

.-. a = 63° 15' 8", 6= 43° 53' 42 
s - a = 27° 29' 17". 
s - 6 = 46° 50' 43". 



•. (a -6) 2 = 9° 40' 43". 



log sin a = 9.95085 -10 
log sin 6 = 9.84094-10 



log sin (s-a)= 9.66423 - 10 

log sin (s - b) = 9.86303 - 10 

.-. log numerator = 19.52726 — 20 

.-. log sin (C/2) = (19.73547 - 20)/2 = 9.86774 
.-. C = (47° 30' 55") • 2 = 95° V 50". 

By the law of sines we obtain C = 95 c 1', using a five-place table 
95° V 20", using a seven-place table. 



.-. log denominator = 9.79179 - 10 
10. 



and 



EXERCISE XLIII 

1. Given a = 56° 40', b = 30° 50', A = 103° 40'; to find B, C, c. 

sin B = sin b sin A/sin a. 
cot (C/2) = sin § (a + b) tan § (A - J5)/sin i (a - b). 
tan (c/2) = sin \{A+B) tan i (a - 6)/sin ± ( A - J?). 

log sin b = 9.70973 - 10 (a + b)/2 = 43° 45'. 

log sin ^L = 9.98753 - 10 (a - &)/2 = 12° 55'. 

.-. log product = 19.69726 - 20 (A + B)/2 = 70° 7' 44". 

log sin a = 9.92194-10 (A - B)/2 = 33° 32' 16" 
.-. log sin £= 9.77532 -10 

.-. B = 36° 35' 28". B < A, since b < a. 



186 



SPHERICAL TRIGONOMETRY 



log sin i (a + 6) = 9.83980 - 10 

log tan \ (A -B) = 9.82140 -10 

.-. log product = 9.66120 - 10 

log sin \ (a - b) = 9.34934 - 10 

.-. log cot (C/2) = 0.31186 

.-. C/2 = 25° 59' 53", 
.: C = 51° 59' 45". 



log sin | (A+B) = 9.97334 - 10 

log tan i (a-b) = 9.36047 - 10 

.-. log product = 9.33381 - 10 

log sin \{A-B) = 9.74232-10 

.-. log tan (c/2) = 9.59149 - 10 

.\-c/2 = 21°19 / 29 // . 
.-. c = 42° 38' 58". 



2. Given b = 100°, c = 62°, B = 95° ; to find a, A, C. 



sin C — sin c sin B/ sin b. 
cot (A/2) = sin § (6 + c) tan £ 



C)/sini(6-c). 



tan (a/2) = sin | (B + C) tan | (b - c)/sin i (5 - C). 
log sin c = 9T. 94593 - 10 (b + c)/2 = 81°. 



9.99834-10 



log sin 5 

\ log product = 19.94427 

log sin b- 9.99335 

.-. log sin C 



20 
10 



(b - c)/2 = 19°. 
(B- C)/2 = 15°51 / 52 / 
(£ + C)/2 = 79° 8' 9". 



9.95092 - 10 
.-. C = 63° 16' 17". C < B, since c < b. 



logsini(& + c)= 9.99462-10 

log tan i (B - C) = 9.45361 - 10 

.-. log product - 19.44823 - 20 

log sin | (6 - c) = 9.51264-10 

.-. log cot (4/2) = 9.93559 - 10 

.-. A/2 = 49° 14'. 
.-. A = 98° 28'. 



log sin i(B+C) = 9.99214 - 10 

log tan i (b - c) = 9.5 3697 -10 

.-. log product = 9.52911 - 10 

log sin \ (B-C) = 9.43674 - 10 

.-. logtan(a/2) = 0.09237 

..-. a/2 = 51° 2' 51". 
.-. a = 102° 5' 42". 



*. Given a = 43° 20', b = 48° 30', A = 58° 40' ; to find B, C, c. 

sin B = sin b sin A/sin a. 
cot (C/2) = sin \ (b + a) tan | (5 - 4)/sin | (b - a). 
tan (c/2) = sin \ (B + A) tan \ (b - a)/sin £ (-B - 4). 

log sin b = 9.87446 - 10 (6 + a)/2 = 45° 55'. 

log sin 4 = 9.93154 - 10 (6 - a)/2 = 2° 35'. 

.-. log product = 9.80600 - 10 (B x + A)/2 = 63° 43' 30' 

log sin a = 9.83648 - 10 (B r - A)/2 = 5° 3' 30". 

.-. log sin B = 9.96952 - 10 (B 2 + A)/2 = 84° 56 / 30' 



B 1 = 68° 47'. 
B 2 =lll°13 / . 



(B 2 - A)/2 = 26° 16' 30' 



EXERCISE XLIII 



187 



Both values of B satisfy condition II in § 120. 



log sin i (6 + a) = 


: 9.85632 - 


10 


logsha<Bi + 4) = 


■ 9.95264 - 


10 


logtani(£i-^i) = 


8.94702 - 


10 


log tan | (6 — a) = 


: 8.65435 - 


10 


.-. log product = 


: 8.80334 - 


10 


.-. log product = 


: 8.60699 - 


10 


log sin |(6 — a) = 


: 8.65391 - 


10 


log sin i (Ex -.4) = 


: 8.94532 - 


10 


.: logcot(Ci/2) = 


0.14943 




.-. log tan (ci/2) = 


- 9.66167 - 


10 


.-. d/2 = 


35° 19' 55 




.-. Cl /2 = 


: 24° 38' 53 


". 


.-. 1 = 


70° 39' 50 


" m 


.-. Ci = 


: 49° 17' 46 


" '. 


log sin i (6 + a) = 


9.85632 - 


10 


logsini(B 2 + ^) = 


: 9.99831 - 


10 


logtani(2?2 — A) = 


: 9.69345 - 


10 


log tan \ (b — a) = 


: 8.65435 - 


10 


.-. log product = 


: 9.54977 - 


10 


.-. log product = 


: 8.65266 - 


10 


log sin l(b — a) - 


: 8.65391 - 


10 


\ogsml(B 2 -A) = 


: 9.64609 - 


10 


.-. logcot(C 2 /2) = 


= 0.89586 




.-. log tan (c 2 /2) = 


: 9.00657 - 


10 


.-. C 2 /2 = 


■ 7° 14' 36" 




.-. c 2 /2 = 


5° 47' 50" 




.-. C 2 = 


14° 29' 12' 


'. 


.-. c 2 = 


: 11° 35' 39 





4. Given a = 41° 6', c = 48° 22', A = 54° 17'; to find 6, B, C. 

sin C = sin c sin A /sin a. 
cot (B/2) = sin £ (c + a) tan i (C - A)/sm i(c- a). 
tan (6/2) = sin £ (C + A) tan J (c - a)/sin $(C -A) 

log sin c 



9.87356 - 10 

log sin A = 9.90951 - 10 

log product = 19.78307 - 20 

log sin a = 9.81781 -10 



log sin C = 9.96526 -10 

.-. d = 67° 23' 12". 
C 2 = 112° 36' 48". 



(c + a)/2 = 44° 44'. 

(c - a)/2 = 3° 38'. 
(Ci - A)/2 = 6° 33' 6". 
(Ci + 4)/2 = 60° 50' 6". 
(C 2 - ^4)/2 = 29° 9' 54". 
(C 2 + ^)/2 = 83°26'54' 



Both values of C satisfy condition II in § 120. 



log sin i(c + a) = 9.84745 - 10 

log tan l(C x -A) = 9.06013 - 10 

.-. log product = 8.90758 - 10 

logsin£(c - a) = 8.80189-10 

.-. logcot(£/2) = 0.10569 



Bi/2 
.-. B x 



38° 5' 46". 
76° 11' 32' 



logsini-(Ci + ^.)= 9.94113-10 

logtan£(c - a) = 8.80277 - 10 

.-. log product = 18.74390 - 20 

logsinj(Ci-^i)= 9.05728-10 
.-. log tan (6i/2) = 9.68662 - 10 



h/2 



25° 55' 8". 
51° 50' 15' 



188 



SPHERICAL TRIGONOMETRY 



log sin £ (c + a) 
logtan£(C 2 -4) 



9.84745-10 
9.74670-10 



.-. log product = 9.59415 - 10 
log sin \{c~a) = 8.80189-10 
.-. logcot(B 2 /2)-= 0.79226 



B 2 /2 = 9° 
.-. B 2 = 18 c 



Y 54". 
19' 48' 



logsini(C 2 + ^L)= 9.99716-10 

log tan i(c- a)= 8.80277 - 10 

.-. log product = 18.79993 - 20 

log sin i(C 2 -A)= 9.68782 -10 
.-. logtan(6 2 /2) = 9.11211 - 10 

.-. b 2 /2 = 7° 22 / 34". 
.-. b 2 = 14° 45' 8". 



5. Given A = 108° 40', B = 134° 20', a = 145° 36'; to find b, C, c. 

sin b = sin B sin a/sin A . 
cot (0/2) = sini(6 + a) tan ±(B - A) /sin $(b - a), 
tan (c/2) = sin i (B ■+ J.) tan $ (6 - a)/sin £ (J3 - 4) . 

log sin 5 = 9.85448 - 10 (6 + a)/2 = 150° 10' 32 ,/ . 

log sin a = 9.75202 -10 (b - a)/2 = 4° 34' 32 // . 

,-. log product = 19.60650 - 20 (B + 4)/2 = 121° 30'. 

log sin A = 9.97653 - 10 (B - A)/2 = 12° 50'. 
.-. log sin b= 9.62997 -10 



.-. b = 154° 45' 4 // . 

log sin £ (6 + a) = 9.69665 - 10 

log tan i(B-A) = 9.35757-10 

.-. log product = 9.05422 - 10 

log sin | (6 - a) = 8.90186-10 
.-. log cot (C/2) = 0.15236 

.-. C/2 = 35° 9'. 
.-. C=70°18'. 



b >a, since B> A. 

log sin i (B + A) = 9.93077-10 

log tan £ (b - a) = 8.90325 -10 

.-. log product = 18.83402 - 20 

log sin i{B- A)= 9.34658 - 10 

.-. log tan (c/2) = 9.48744 - 10 



.-. c/2 
.-. c 



17° 4' 40" 
34° 9' 20' 



6. Given B = 116°, C = 82°, c = 86° ; to find a, 4, b. 

sin & = sin c sin S/sin C. 
cot (4/2) = sin i (& + c) tan £ (B - C)/sin i (& - c). 
tan (a/2) = sin $(B+C) tan £ (& - c)/sin £ (£ - C). 

log sin c = 9.99894 - 10 (B + C)/2 = 99°. 

log sin B= 9.95366-10 (JS - C)/2 = 17°. 

.-. log product = 19.95260 - 20 (b + c)/2 = 100° 33' 35' 

log sin C = 9.99575-10 (6 - c)/2 = 14° 33' 35". 
.-. log sin b= 9.95685 -10 

.-. 6 = 115° 7' 10". b > c, since B > C. 



EXERCISE XLIII 



189 



log sin £ (b + c) = 9.99258 - 10 

logtan£(£- C)= 9.48534-10 

.-. log product = 19.47792 - 20 

log sin £ (b - c) = 9.40085 - 10 

.-. log cot (A/2) = 10.07757 - 10 

.-. ^4/2 = 39° 54' 37". 
.-. A = 79° 49' 14". 



log sin £ (B + C) = 9.99462-10 

log tan £ (b - c) = 9.41452 - 10 

.-. log product = 19.40914 - 20 

log sin % (B - C) = 9.46594 - 10 
.-. log tan (a/2) = 9.94320 - 10 

.-. a/2 = 41° 15' 50". 
.-. a = 82° 31' 40". 



7. Given A = 121°, B = 108°, a = 130°; to find 6, C, c. 

sin 6 = sin a • sin .B/sin ^1. 
cot ( C/2) = sin £ (a + 6) tan £ (J. - J3)/sin £ (a - 6) . 
tan (c/2) = sin £ (A + B) tan £ (a - 6)/sin i(A- B). 



log sin a = 9.88425-10 
log sin B = 9.97821 - 10 

log product = 19.86246 - 20 
log sin A = 9.93307 - 10 

.-. log sin 6= 9.92939-10 

.-. &J. = 58° 12' 23". 
& 2 = 121° 47' 37" 



(A + B)/2 = 114° 30'. 
(4 - B)/2 = 6° 30'. 
(a + &i)/2=94°6 / 11". 
(a - &i)/2 = 35° 53' 49". 
(a + b 2 )/2 = 125° 53' 49' 
(a - & 2 )/2 = 4° 6' 11". 



Since 5 < A, both values of b satisfy condition II in § 120. 



log sin J (a + &i) = 9.99889 - 10 

log tan i(A- B)= 9.05666 - 10 

.-. log product = 19.05555 - 20 

log sin £ (a - &i) = 9.76814 -10 

log cot (Ci/2) = 9.28741 - 10 



.-. d/2 
.-. d 

log sin £ (a + h) 

log tan i(A- B) 

.-. log product 

log sin £ (a — 62) 

.-. logcot(C 2 /2) 

.-. C 2 /2 
.-. C, 



79° V 50" 
158° 3' 40' 

9.90853- 
9.05666 - 



10 
10 



.96519-10 
.85461-10 



0.11058 

37° 47'. 
75° 34'. 



log sin i{A + B) 

log tan £ (a — 61) 

.-. log product 

\og&m\(A-B) 

.-. logtan( Cl /2) 

••• ci/2 



9.95902-10 
9.85962-10 
9.81864 - 10 
9.05386-10 
0.76478 

80° 14' 51". 
160° 29' 41". 



logsin£(^ + B)= 9.95902-10 

log tan £ (a - & 2 ) = 8.85573-10 

.-. log product = 18.81475-20 

logsin£(^-B) = 9.05386-10 

log tan (c 2 /2) = 9.76089 - 10 



c 2 /2 
.-. c 2 



29° 58' 6". 
59° 56' 12' 



/- 



190 



SPHERICAL TRIGONOMETRY 



8. Given B = 36° 20', C = 46° 30', b = 42° 12' ; to find A, a, c. 

sin c = sin b sin C/sin B. 
cot (4/2) = sin £ (c + &) tan i(C - 5) /sin £(c - &). 
tan (a/2) = sin | (0 + 5) tan i (c - &)/sin i (<7 - 5). 



log sin b = 


9.82719 - 10 


(C + B)/2 = 


41° 25'. 




log sin C = 


9.86056 - 10 


(C-B)/2 = 


5° 5'. 




.-. log product = 


19.68775 - 20 


(ex - 6)/2 = 


6° 33' 41". 




log sin B = 


9.77268 - 10 


( Cl + b)/2 = 


48° 45' 41" 




.-. log sin c = 


9.91507 - 10 


(c 2 -b)/2 = 


41° 14' 19". 




.-. Ci = 


55° 19' 23 // . 


(c 2 + b)/2 -.= 


83° 26' 19". 




c 2 = 


124° 40' 37 // . 








Both values of c satisfy condition II 


in § 120. 






logsini(ci+6) = 


9.87621 - 10 


log sin %(C + B): 


= 9.82055- 


■10 


log tan i (<?-£) = 


8.94917 - 10 


log tan \ (ex — b) ■ 


= 9.06078 - 


10 


.-. log product = 


18.82538 - 20 


.-. log product : 


= 18.88133 - 


■20 


log sin i (d - b) = 


9.05792 - 10 


log sin i (C-B): 


= 8.94746- 


10 


.-. log cot (A/2) = 


9.76746 - 10 


.-. log tan (a/2) : 


= 9.93387- 


10 


.: A x /2 = 


59° 39' 17". 


.-. ax/2 : 


= 40° 39' 16" 




.-. 4i = 


119° 18' 33". 


.-. Oi: 


= 81° 18' 32" 




log sin i (c 2 + 6) = 


9.99715 - 10 


log sin I (C + B)-. 


= 9.82055 - 


•10 


logtani(C- 75) = 


8.94917 - 10 


log tan i (c 2 — b) ■. 


= 9.94281 - 


10 


.-. log product = 


18.94632 - 20 


.-. log product : 


= 19.76336 - 


•20 


log sin i (c 2 — 6) = 


9.81911 - 10 


log sin % (C-B)-. 


= 8.94746 - 


10 


.-. log cot A 2 /2 = 


9.12731 - 10* 


.-. log tan (a 2 /2) : 


= 0.81590 




.-. A 2 /2 = 


: 82° 21' 51" + . 


.-. a 2 /2 


= 81° 18' 46' 


'+. 


.-. A 2 = 


164° 43' 43". 


.-. a 2 


= 162° 37' 33 


;". 



Given a = 90°, b = 123° 56', A = 52° 43' 44"; to find B, C, c. 

sin B — sin A sin 6/sin a. 
cot (C/2) = sin i (b + a) tan i (B - A) /sin i (6 - a). 
tan (c/2) = sin \ (B + A) tan | (b - a) /sin i (B - A). 

log sin A = 9.90079 - 10 (b + a)/2 = 106° 58'. 

log sin 6 = 9.91891 - 10 (b - a)/2 = 16° 58'. 

. log product = 9.81970 - 10 (B -f 4)/2 = 95° 42' 20' 

log sin a = 0.00000 (B - A)/2 = 42° 58' 36' 
.-. log sin B =9.81970 - 10 

.-. B = 138° 40' 56". B > A, since b > a. 



EXERCISE XLIII 



191 



log sin I (6 + a) = 9.98067 - 10 

log tan ({B -A) = 9.96930 - 10 

.-. log product = 9.94997 - 10 

log sin | (b- a) = 9.46511 - 10 



log cot (C/2) = 0.48486 
.-. C/2 = 18° V 51". 
.-. C = 36° 15' 42". 



log sin ^(£ + ^4) = 9.99784-10 

log tan i (6 - a) = 9.48443 - 10 

.-. log product = 19.48227 - 20 

log sin | (B - A) = 9.83359 - 10 

.-. log tan (c/2) = 9.64868 - 10 

.-. c/2 = 24° 0' 18". 

.-. c = 48° 0' 36". 



10. Given a = 49° 14. 6', 6 = 41° 9.8', C = 76° 18. 6' ; to find A, B, c. 
tan J (A + B) = cos | (a - b) cot (C/2)/cos |"(a + b). 
tan i (^4 - B) = sin | (a - b) cot (C/2)/sin \ (a + 6). 



sin c = sin a sin C/sin ^4. 
(a + 6)/2 = 45° 12.2', (a - 6)/2 = 4° 2.4', C/2 



log cos I (a - 6) = 9. 99892 - 10 

log cot (C/2) = 0.10477 

.-. log product = 10. 10369 - 10 

log cos \ (a + b) = 9.84794 - 10 

log tan \ (A +B)= 0.25575 

.-. l(A + B) = 60° 58' 20". 

.-. A = 68° 10' 29", B 

log sin a = 9.87938-10 

log sin C = 9.98748 - 10 

.-. log product = 19.86686 - 20 

log sin A = 9.96769 - 10 

.-. log sin c = 

.-. c = 



38° 9.3'. 
8.84790 - 10 
0.10477 



log sin \ (a — 6) 

log cot (C/2) 

.-. log product = 18.95267 - 20 

log sin 1 (a + b) = 9.85102 - 10 

.-. log tan \(A-B) = 9.10165 - 10 

.-. l(A -B) = 7° 12' 9". 
53° 46' 11". 



10 



9.89917 
52° 26' 54". 

Only the less value of c satisfies the condition c < a + 6, or IV in § 120. 
11. Given b = 41° 9.8', C = 32° 20.2', A = 68° 10' 29" ; to find a, B, c. 
tan£(a + c ) = cos£(J. — C)tan(6/2)/cos-£-(^4 + C). 
tan i (a - c) = sin i(A-C) tan (6/2)/sin i (A + C). 
sin 2? = sin b sin ^4 /sin a. 
(4 - C)/2 = 17° 55' 9", (4 + C)/2 = 50° 15' 21", 6/2 = 20° 34' 54". 



log cos 1 (J. -C) = 9.97840-10 

log tan (6/2) = 9.57462 -10 

.-. log product = 19.55302 - 20 

logcosj(^ + C) = 9.80575-10 

logtan£(a + c) = 9.74727-10 

.-. (a + c)/2 = 29° 11' 50". 

.-. a = 37° 44' 33", c : 



log sin i(A-C) 
log tan (6/2) 



9.48809 - 10 
9.57462 -10 



.-. log product = 19.06271 - 20 

log sin l(A+C)= 9.88587 -10 

.-. log tan i (a - c) = 9.17684-10 

.-. (a - c)/2 = 8° 32' 43". 
20° 39' 7". 



192 SPHERICAL TRIGONOMETRY 

log sin b = 9.81836 -10 

log sin A = 9,96769 - 10 

.-. log product = 19.78605 - 20 

log sin a = 9.78683-10 

.-. log sin B = 9.99922 -10 

.-. B = 86° 34'. 

To check B and to determine in what quadrant B is, use the formula 
sin (B/2) in § 116. 

12. Given a = 60°, c = 73°, B = 125° 40' ; to find A, C, b. 

tan % (C + A) = cos £ (c — a) cot (B/2) /cos £ (c + a). 
tan \ (C - A) = sin \{c - a) cot (£/2)/sin £(c + a), 
sin b — sin a sin J5/sin A. 
(c - a)/2 = 6° 30', (c + a)/2 = 66° 30', 5/2 = 62° 50'. 
log cos i (c - a) = 9.99720-10 log sin £ (c - a) = 9.05386-10 

log cot (B/2) = 9.71028-10 log cot (B/2) = 9.71028 -10 

.-. log product = 9.70748 - 10 .-. log product = 18.76414 - 20 

log cos (c + a)/2 = 9.60070 - 10 log sin £ (c + a) = 9.96240-10 

.-. logtan£(C+^) = 0.10678 .-. logtani(C-^i) = 8.80174-10 

.-. (C + ^)/2 = 51° 58' 25". .-. (C - A)/2 = 3° 37' 29". 

.-. A = 48° 20' 56", C = 55° 35' 54". 
log sin a = 9.93753-10 
log sin B = 9.90978-10 
.-. log product = 19.84731 - 20 
log sin A = 9.87344-10 
.;. log sin b= 9.97387 -10 
.-. b = 109° 40' 36". b > c, since B > C. 

13. Given a = 150° 57.1', b = 134° 15.2', A = 144° 22.3'; to find B, C, c. 

sin B = sin b sin A /sin a. 
cot(C/2) = sin£(a + b) tan £ ( A - B)/sin £ (a - b). 
tan (c/2) = sin \{A+B) tan £ (a — b) /sin | (^4 - JB). 
log sin 6 = 9.85507 - 10 (a + 6)/2 = 142° 36' 9". 

log sin ^4 = 9.76532 -10 (a - b)/2 = 8° 20' 57". 

.-. log product = 19.62039 - 20 (A + B x )/2 = 101° 48' 24". 

log sin a = 9.68623 - 10 (A - B x )/2 = 42° 33' 54". 

.-. log sin B = 9.93416 - 10 (A + B 2 )/2 = 132° 33' 54". 

.-. B 1 = 59° 14' 30". (A ~ 30/2 - H° 48' 24". 

£ 2 = 120° 45' 30". 



EXERCISE XLIII 



193 



Both these values 

log sin £ (a + b) = 

log tan £ (A — B\) = 

.-. log product = 

log sin i (a — b) = 

.-. logcot(Ci/2) = 

a (Ci/2) = 

.-. d = 

log sin i (a + b) = 

log tan %(A — B 2 ) = 

.-. log product = 

log sin $■-(«—&) = 

.-. logcot(C 2 /2) = 

■•• C 2 /2 = 

.-. C 2 = 



of B satisfy condition II in § 120. 



9.78343-10 

9.96304-10 

9.74647-10 

9.16199-10 

0.58448 

14° 35' 30". 

29° 11'. 
9.78343 - 10 
9.32021 - 10 

19.10364 -20 
9.16199 -10 
9.94165- 10 

48° 50' 14". 

97° 40' 28". 



log sin i (A + Bi) 

log tan £ (a - b) 

.-. log product 

log sin \{A- Bi) 

.-. logtan(ci/2) 

'.-. Cl /2 

.-. Ci 

log sin i{A + B 2 ) 

log tan ■§• (a — b) 

.-. log product 

\ogsini(A-B 2 ) 

.-. log tan (c 2 /2) 

.-. c 2 /2 

.-. c 2 



9.99071 - 10 

9.16661-10 

9.15732-10 

9.83022-10 

9.32710-10 

11° 59' 24". 

23° 58' 48". 

9.86718-10 

9.16661-10 



9.03379 
9.31092 



9.72287- 10 
27° 50' 48". 
55° 41' 36". 



14. Given A = 1 
sin b 
cot (0/2) 

tan (c/2) 
log sin a 
log sin B 
.-. log product 
log sin A 
.-. log sin b 



00° 2.4', B = 98° 30.5', a = 95° 20.6' ; to find 6, c, C. 

= sin a sin 2?/sin J.. 

= sin $ (b + a) tan } (.4 - B)/sin £ (a - 6). 

= sin £ (4 + 5) tan £ (a - 6) /sin J (.4 -' 5). 

= 9.99811 - 10 (J. + B)/2 = 99° 16' 27". 

= 9.99519-10 • (A - B)/2 = 45' 57". 

= 19.99330 - 20 (6 + a)/2 = 92° 40' 18". 

= 9.99330 - 10 (a - 6)/2 = 2° 40' 18". 



b = 



log sin i(b + a) 

log tan | ( J. - B) 

.-. log product 

log sin i(a — b) 

.: log cot (C/2) 

.-. C/2 

.-. C 



0.00000 
90°. 

9.99953 
8.12603 



18.12556 -20 
8.66850 - 10 



9.45706 - 

74° 0' 55" 
148° 1' 50' 



10 log sin i (A + B) = 9. 99428 -10 

JL0 log tan }(<* — &)= 8.66897 -10 

.-. log product = 8.66325 - 10 
log sin i(A-B) = 8.12600-10 
10 .-. log tan c/2 = 0.53725 

.-. c/2 = 73° 48' 55". 
.-. c = 147° 37' 50". 



15. Given a = 131° 35.7', & = 108° 30.9', c = 84° 46.8' ; to find A, B, C. 

tan r — Vsin (s — a) sin (s — b) sin (s — c)/sin s. 
tan (A/2) = tan r/sin (s ~ a), etc. 

2 s = 324° 53.4'. .-. s = 162° 26.7', 
s - a = 30° 51', s - b = 53° 55.8', s - c .= 77° 39. 9'. 



194 SPHERICAL TRIGONOMETRY 

log sin (s- a)= 9.70994 - 10 .-. log tan (A/2) = 0.35402. 

log sin (s - 6) = 9. 90757 - 10 log tan (B/2) = 0. 15639. 

log sin (s-c) = 9.98986 - 10 log tan (C/2) = 0.07410. 

.-. log product = 19.60737 - 20 .*. A/2 = 66° V 39". 

log sin s = 9.47946 - 10 B/2 = 55° 6'. 

.-. log tan r = 0.12791/2 C/2 = 49° 51' 53". 
= 10.06396-10. 
.-. A = 132° 15' 18", B = 110° 12', C = 99° 43' 46". 

16. Given ^1 = 4° 23.4', 5 = 8° 28. 7', C = 172° 17. 6'; to find a, 6, c. 

tan R = V- cos S/[cos (S - A) cos (S - B) cos (S - C)]. 
tan (a/2) = tan B cos (S — A), etc. 
S = 92° 34' 51", 8 -^ = 88°ll / 27 ,/ , .S - JB= 84° 6' 9", S-C= -79°42 / 45 ,/ . 
log cos (5 -A) = 8.49928 - 10 .-. log tan (a/2) = 9.44457 - 10. 

log cos (S - B) •= 9.01178 - 10 log tan (6/2) = 9.95707 - 10. 

log cos (S-C) = 9.25185-10 log tan (c/2) = 0. 19714. 

.'•. log product = 6.76291 - 10 .-. a/2 = 15° 33' 13"+. 

log cos S = 8.65349 - 10 6/2 = 42° 10" 22". 

.-. log tan B = 1.89058/2 c/2 = 57° 34' 45". 

= 0.94529. 
.-.. a = 31° 6 7 27", 6 = 84° 20' 43", c = 115° 9' 30". 
In the following examples p will denote that perpendicular from the 
vertex C to the side AB, whose foot is towards B from A. 

17. If 6 = 90°, prove that p = A. 

The great circle whose pole is the vertex A will pass through C and be 
perpendicular to AB ; hence the arc between C and AB, or AB produced 
through JB, will measure A. But this arc is p ; hence p — A. 

18. If 6 = 90° and A < 90° ; discuss the case when a < A, a = A, 
a> A and < 6, a = 6, a > 6. See V in § 106. 

Since 6 = 90°, p = A ; hence since A < 90°, p < 90°. 

Hence p is the shortest arc from C to AB by (i) in § 104. 

When a < A, a <p; hence no triangle is possible. 

When a = A, a = p; hence the triangle is right-angled at B. 

When a> A and < 6, a > p and < 6 ; also a > p and < 180° — 6 ; hence 
there are two triangles having the given parts. 

When a = 6, the triangle becomes the arc CA , or the lune whose angle 
is^L°. 

When a > 6, no convex triangle is possible. 






EXERCISE XLIII 195 

19. If 6 = 90° and ^4>90°; discuss the case when a>A, a — A, 
a < A and > b, a = 6, a < b. 

Since b = 90°, p = A; hence since A > 90°, p > 90°. 

Hence p is the longest arc from C to .AS by (i) in § 104. 

When a> A, a> p; hence no triangle is possible. 

When a = A, a = p ; hence the triangle is right-angled at B. 

When a < A and >b, a<p and > 6 ; also a < p and > 180° — b ; hence 
there are two triangles having the given parts. • 

When a = b, the triangle becomes the arc CA or the lune whose angle 
is^.°. 

When a < b, no convex triangle is possible. 

20. If b = 90° and A = 90° ; discuss the case when a-^ A, a— A. 
When A = 90° and b = 90°, C is the pole of arc AB, and any arc of a 

great circle from C to AB is perpendicular to AB and equal to 90°. 

Hence when a ^ A, a ^ 90°, and no triangle is possible. 

When a = A, a = 90° ; hence B = 90°, but C and c are indeterminate, 
and therefore the triangle is indeterminate. 

21. If A = 90° and b < 90° ; discuss the case when a < b, a = 6, 
a > b and < 180° - 6, a = 180° - 6, a > 180° - 6. 

When A = 90°, .4C ± y!£ and p = b. 
When a < 6, a < _p ; hence no triangle is possible. 
When a — 6, the triangle becomes the arc A C. 

When a > b and < 180° — 6, a > p and < 180° — b ; hence there is one 
convex triangle. 

When a = 180° — 6, the triangle becomes the lune whose angle is A°. 
When a > 180° — 6, no triangle is possible. 

22. If A < 90° and b < 90° ; discuss the case when a<p, a = p, 
a>p and < 6, a = 6, a > 6 and < 180° - 6, a = 180° - 6, a > 180° - 6. 

When J. < 90° and b < 90°, p < b. 

When a < p, no triangle is possible. 

When a = p, the triangle is right-angled at U. 

When a> p and < 6, we have also a>p and < 180° — b ; hence there 
are two triangles having the given parts. 

When a = 6, a>p and < 180° — b ; hence there is one triangle. 

When a > b and < 180° -6, a>p and < 180° - b ; hence there is one 
triangle. 

When a = 180° — 6, the triangle becomes the lune whose angle is A . 

When a > 180° — 6, no convex triangle is possible. 



196 SPHERICAL TRIGONOMETRY 

23. If A > 90° and b < 90° ; discuss the case when a > p, a = p, 
a<p and > b or 180° -b,a = b,a>b and < 180° - 6, a > 180° - & and < p. 

When J. > 90°, p > 6 ,or 180° - 6. 

When a >_p, no triangle is possible. 

When a = p, the triangle is right-angled at B. 

When a < p and > b or 180° — b, there are two triangles. 

When a = 6, the triangle becomes the arc AC. 

When- a > 6 and < 180° — 6, there is one triangle, as then a lies between 
b and p, but not between p and 180° — &. 

When a > 180° — b and < jp, there is one triangle, as in this case a 
lies between p and 180° — 6, but not between p and b. 

EXERCISE XLIV 

1. Find the area of a spherical degree and of a lune of angle 43° 30' on 
a sphere whose radius is 9 ft. 

Spherical degree = 7rr 2 /180 = tt9 2 /180 = 1.414 sq. ft. 

Lune of angle 43° 30' = 87 spherical degrees = 87 x 1.414 = 123 sq. ft. 

2. Find the area between the meridian 80° W. and 87° W., the radius 
of the earth being 3960 miles. 

Spherical degree = it ■ 3960 2 /180 - 273,696. 

.-. lune of angle 7° = 14 x 273,696 = 3,831,744 sq. mi. 

3. Find the area of the triangle ABC when on a sphere whose radius 
is 10 yd., having given : 

Ex. 1. a = 58° 18', b = 62° 40', c = 78° 24'. 

tan (E°/4) — Vtan (s/2) tan i (s — a) tan |(s — 6) tan ±(s- c). 
s/2 = 49° 50' 30", 1 (s - a)= 20° 41' 30", 
i (s - b) = 18° 30' 30", \ (s - c) = 10° 38' 30". 

log tan 49° 50' 30" = 0.07375 
log tan 20° 41' 30" = 9. 57715 - 10 
log tan 18° 30' 30" = 9. 52473 - 10 
log tan 10° 38' 30" = 9.27392 - 10 
.-. log tan (E°/4) =(18.44955 - 20)/ 2 
= 9.22478 - 10 

.-. E°/4: = 9° 31' 31". .-. E° = 38° 0' 3" = 38.101°. 

An „ 38.101 • Tt ■ 10 2 38.101x5.236 aa ^ , ■ 

.-. Area ABC = = = 66.5 sq. yd. 

180 3 



EXERCISE XLIV 197 

Ex. 2. a = 110° 4', 6 = 74° 32', c = 56° 30'. 

tan (E°/i) = Vtan (s/2) tan i (s — a) tan J (s — 6) tan \ (s — c). 

s/2 = 60° 16' 30", i ( s - «) = 5 ° 14 ' 3Q// » 
' i (s - 6) = 23° 0' 30", i (s-c) = 32° 1' 30". 

log tan 60° 16' 30" = 0. 24339 
log tan 5° 14' 30" = 8.96256 - 10 
log tan 23° 0' 30" = 9. 62803 - 10 
log tan 32° r 30" = 9.79621-10 
.-. log tan [E°/i) = (18.63019 - 20)/2 
9.31510-10 

.-. j£°/4 = 11° 40' 20". .-. E° = 46° 41' 20" = 46.689°. 

A1>r , 46.689- 7t- 102 155.63 -it 

.-. Area ABC = = = 81.49 sq. yd. 

180 6 * J 

Ex. 3. A = 53° 50', B = 96° 18', C = 64° 20'. 

£° = 53° 50' + 96° 18' + 64° 20' -180°= 34° 28' = 34. 4f °. 

AJ}n 34.4f.7T. 102 172.31- 7T aa ^ ro 

.-. Area ABC = ? = 2 = 60.156 sq. yd. 

180 9 j J 

Ex. 4. A = 74° 40', J5 = 67° 30', C = 49° 50'. 

E° = 74° 40' + 67° 30' + 49° 50' - 180° = 12°. 

., Area ABC = 12 -*- 102 = ™1 = 20.94 sq. yd. 
180 3 

4. Find the area of the triangle ABC, when on a sphere whose radius 
is 100 ft. , having given : 

Ex. 1. a = 64° 24', 6 = 42° 30', C = 58° 40'. 

E° _ tan (a/2) tan (6/2) sin C 
tan — — 



2 1 + tan (a/2) tan (6/2) cos C 



log tan 32° 12' = 9. 79916 - 10 log tan 32° 12' = 9. 79916 - 10 

log tan 21° 15' = 9.58981 - 10 log tan 21° 15' = 9.58981 - 10 

log sin 58° 40' = 9.93154 - 10 log cos 58° 40' = 9.71602 -10 

.-. log numerator = 9.32051 - 10 .-. log product = 9.10499 - 10 

los: denominator = 0.05206 



.-. product = 0.87265. 
log tan (E°/2) = 9.26845 - 10 ... denominator = 1.87265. 

.-. E°/2 = 10° 30' 41", or E° = 21° V 22" = 21.228°. 
.-. Area ABC = 21.228 % 100 2 /180 = 70.76 x 52.36 = 3705 sq. ft. 



198 SPHERICAL TRIGONOMETRY 

Ex. 2. c = 78° 15', b = 56° 20', A = 120°. 

E° _ tan (6/2) tan (c/2) sin ^4 
2 _ 1 + tan (6/2) tan (c/2) cos A ' 

log tan 39° 7' 30" = 9.91031 - 10 log tan 39° V 30" = 9.91031 - 10 

log tan 28° 10' = 9.72872-10 log tan 28° 10' = 9.72872 - 10 

log sin 120° = 9.93753 - 10 log cos 120° = 9.69897 - 10 

.-. log numerator = 19.57656 - 20 .-, log product = 9.33800 - 10 

log denominator = 9.89335 — 10 . product = — .21777. 

.-. log tan (E°/2) = 9.68321 - 10 ... denominator = 0.78223. 

.-. E°/2 = 25° 44/ 33", or E° = 51° 29' 6" = 51.485°. 
.-. Area ABC = 51.485 it 100 2 /180 = 51.485 x 523.6/3 = 8985.8 sq. ft. 

Ex. 4. B = 98° 30', C = 67° 20', a = 60° 40'. 

By example 4 in Exercise XLII, we have A = 59° 43' 53". 

.-. E° = 98° 30' + 67° 20 7 + 59° 43' 53" - 180 = 45.565°. 
.-. Area ABC = 45.565 it 100 2 /180 = 45.565 x 523.6/3 = 7952.6 sq. ft. 

Ex. 5. A = 125° 20', G = 48° 30', 6 = 83° 13'. 

By example 5 in Exercise XLII, we have B = 62° 54 y 34". 

.-. E° = 125° 20' + 48° 30' + 62° 54' 34" - 180° = 56.743°. 
..-. Area ABC = 56.743 it 100 2 /180 = 56.743x523.6/3 = 9903.5 sq. ft. 

5. Find the area of the triangle ABC, when on a sphere whose radius 
is 3960 mi., having given : 

Ex. 1. a = 56° 40', 6 = 30° 50', A = 103° 40'. 

By example 1 in Exercise XLIII, we have 

B = 36° 35' 29", C = 51° 59' 45". 
.-. E° = 103° 40' + 36° 35' 28" + 51° 59' 45" - 180° = 12.254°. 
.-. Area ABC = 12.254 it 3960 2 /180 = 12.254 x 396 x 132 x 5.236 

= 3,354,000 sq. mi. (by logs.) 

Ex. 2. b = 100°, c = 62°, B = 95°. 

By example 2 in Exercise XLIII, we have 

A =98° 28', = 63° 16' 17". 
.-. E° = 95° + 98° 28' + 63° 16' 17" - 180° - 76.738°. 
.-. Area ABC = 76.738 7t3960 2 /180 = 76.738 x 396 x 132 x 5.236 

= 21,003,000 sq. mi. (by logs.) 



EXERCISE XLV 199 

Ex. 5. A = 108° 40', B = 134° 20', a = 145° 36'. 

By example 5 in Exercise XLIII, we have C = 70° 18'. 

.-. E° = 108° 40' + 134° 20' + 70° 18' - 180 = 133.3°. 
.-. Area ABC = 133.3 n 39607180 = 133.3 x 396 x 132 x 5.236 

= 36,484,000 sq. mi. (by logs.) 

Ex. 6. C = 82°, B = 116°, c = 86°. 

By example 6 in Exercise XLIII, we have A = 79° 49' 14". 
.-. E° = 82° + 116° + 79° 49' 14" - 180° = 97.821°. 
.-. AvezABC = 97.821 #39607180 = 97.821 x 396 x 132 x 5.236 

= 26,773,000 sq. mi. (by logs.) 

EXERCISE XLY 

1. Eind the shortest distance, measured on the earth's surface, between 
Boston (42° 21' N., 71° 3' W.) and Cape Town (33° 56' S., 18° 28' E.), and 
the bearing of each city from the other. 

For the solution, see the text in Exercise XLIV and § 123. 
BC = 111° 46' 24" = 111.771°. 
1° = (2 it • 3960/360) mi. = 22 it mi. 
.-. BC = 111.771 x 22 7T mi. = 7725.2 mi. 

2. Find the longitude of B, the place where BC, in fig. 79, crosses the 
equator, the distance RC, and the bearing of Cape Town from R. 

In the spherical triangle NRC, we have 

RN = 90°, CN = 123° 56', Z NCR = 52° 43' 44". 
By example 9 in Exercise XLIII, we have 

Z CRN = 138° 40' 56", RC = 48° 0' 36", Z RNC = 36° 15' 42". 

Hence the bearing of Cape Town from R is S. (180° - Z CRN) E., or 
S. 41° 19' 4" E. 
The distance RC = 48.01° = 48.01 x 22 n mi. = 3318.2 mi. 
The longitude of R = (Z RNC - Z GNC)W. 

= (36° 15' 42" - 18° 28')W. = 17° 47' 42" W. 

3. Find the shortest distance between New York (40° 45.4' N., 73° 
58.4' W.) and Paris (48° 50.2' X., 2° 20.2' E.), and the bearing of each 
city from the other. 

Let ABC be the triangle whose vertices are New York (B), Paris (A), 
and the North Pole (C). 



200 SPHERICAL TRIGONOMETRY 

Then a = 90° - 40° 45.4' = 49° 14.6', 

b = 90° - 48° 50. 2' = 41° 9.8'; 
C = 73° 58.4' + 2° 20.2' = 76° 18. 6'. 

By example 10 in Exercise XLIII, we have 

A = 68° 10' 29", B = 53° 46' 11", c = 52° 26' 54". 

Hence the bearing of Paris from New York is N. 53° 46" 11" E., and 
the bearing of New York from Paris is N. 68° 10' 29" W. ; the distance 
BA = 52.45° = 52.45 x 22 it mi. = 3625 mi. 

4. A ship is sailing westward on a great circle which passes through 
New York and Paris ; find its latitude when it crosses the meridian 30° W. ; 
find also its course and its distance from Paris. 

Let ABC be the triangle whose vertex A is Paris, whose vertex C is 
the North Pole, whose side CB is an arc of the meridian 30° W., and 
whose side A B is an arc of the great circle through Paris and New York ; 

then b = 90° - 48° 50.2' = 41° 9.8', 

C = 30° + 2° 20.2' = 32° 20.2', 
A = 68° 10' 29". 

By example 11 in Exercise XLIII, we have 

a = 37° 44' 33", B = 86° 34', c = 20° 39' 7". 

Hence the latitude = 90° - 37° 44' 33" = 52° 15' 27"; the course is S. 
86° 34' W. ; the distance = 20.65$° = 20.651 x 22 it mi. = 1427.3 mi. 

5. From a place 17° N. 130° W. a ship, starting on a course S. 
54° 20' W. , sailed on a great circle ; find its latitude and longitude when 
it has sailed 60°, or 4146.9 mi. ; find also its course. 

Let ABC be the triangle whose vertex B is the place whence the ship 
sails, whose vertex A is the North Pole, and whose side BC, or a, is the 
distance sailed ; then 

a = 60°, c = 73°, B = 125° 40'. 

By example 12 in Exercise XLIII, we have 

A = 48° 20' 56", C = 55° 35' 54", b = 109° 40' 36". 

Hence the longitude of the ship = (130° + 48° 20' 56") W. = 178° 20" 56" W. ; 
its latitude = (109° 40' 36" - 90°) S. = 19° 40' 36" S. ; and its course is S. 
55° 35' 54" W. 



EXERCISE XLV 201 

6. A ship sails from San Francisco (37° 47' 55" N, 122° 24' 32" W.) 
on a great circle to Yokohama (35° 26' N., 139° 39' E.) ; find the distance 
sailed, and the course of the ship when leaving San Francisco and when 
approaching Yokohama. 

In A ABC let C be San Francisco, B Yokohama, and A the North 
Pole ; then a = 97° 56' 28", 6 = 52° 12' 5", c = 54° 34'. 
By example 3 in Exercise XLII, we have 

a = 74° 33', B = 54° 17' 6", C = 56° 50' 54". 

Hence Yokohama is N. 56° 50' 54" W. from San Francisco, San Fran- 
cisco is N. 54° 17' 6" E. from Yokohama, and the distance between them 
is 74.55° = 74.55 x 22 it mi. = 5152.5 mi. 

7. In example 6 find the distance sailed and the latitude and the course 
of the ship when it crosses the meridian 180°. 

In A ABC let the vertex C be San Francisco, A the North Pole, CB 
or a an arc of the great circle through San Francisco and Yokohama, 
and AB or c an arc of the meridian 180° ; then 

A = 57° 35' 28", 6 = 52° 12' 5", C = 56° 50' 54". 
To find a, B, c, we have the formulas 

tan \(a-c) = sin I {A - C) tan (6/2)/sin \ (A + C), 
tan \ (a + c) — cos \ (A — C) tan (&/2)/cos §(A+C), 
sin B = sin b sin ^4/sin a. 
(A - C)/2 = 22' 17", {A + C)/2 = 57° 13' 11", 6/2 = 26° 6' 3". 
log sin-i (A-C)= 7.81170 - 10 log cos ±{A -C) = 9.99999 - 10 

log tan (6/2) = 9.69011 - 10 log tan (6/2) = 9.69011 - 10 

.-. log product = 7.50181 - 10 .-. log product = 9.69010 - 10 

log sin I (A + C) = 9.92466 - 10 log cos \{A+C)= 9.73353 - 10 

.-. log tan i (a - c) = 7.57715 - 10 .-. log tan \ (a + c) = 9.95657 - 10 
.-. (a - c)/2 = 12' 59". .-. (a + c)/2 = 42° 8' 24". 

.-. a = 42° 21' 23", c = 41° 55' 25". 
log sin 6 = 9.89772 - 10 
log sin A = 9.92647 - 10 
.-. log product = 9.82419 - 10 
log sin a = 9.82849 - 10 
.-. log sin B= 9.99570- 10 



B= 81 



o n 



Oi 



Hence distance = 42.35f° = 42.35| x 22 it mi. = 2927.5 mi., latitude 
= (90° - 41° 55' 25") N. = 48° 4' 35" N, course is S. 81° 57' W. 



</ 



202 SPHERICAL TRIGONOMETRY 

8. A ship sailed on a great circle from San Francisco to Sydney 
(33° 52' S., 151° 13' E.) ; find the distance sailed and the course of the 
ship when leaving San Francisco and when approaching Sydney. 

In A SNF let the vertex F be San Francisco, S Sydney, and N the 
North Pole ; then 

s = 90° - 37° 47' 55" = 52° 12' 5", 
/= 90° + 33° 52" = 123° 52', 
N = 360° - (151° 13' + 122° 24' 32") = 86° 22' 28". 

To find F, S, and n, we have the formulas 

tan l(F-S) = sin \ ■(/ - s) cot (JV/2)/sin | (/ + s), 
tan i(F+S) = cos I (/ - s) cot (JV/2)/cos \(f + s), 
sin n = sin 2V sin s/sin S. 

(/ - s)/2 = 35° 49' 58", (/ + s)/2 = 88° 2' 3", #/2 = 43° 11' 14". 

log sin i (/ - s) ■= 9.76746 - 10 log cos | (/ - s) = 9.90887 - 10 

log cot (N/2) = 0.02750 log cot (N/2) = 0.02750 

.-. log product = 9.79496 - 10 .-, log product = 9.93637 - 10 

log sin \ '(/ + s) = 9.99974 - 10 log cos \ (/ + s) = 8.53534 - 10 

.-. logtani(F- S) = 9.79522 - 10 .-. logtani(F+ S) = 1.40103 

.-. (F - S)/2 = 31° 57' 58". .-. (F + S)/2 = 87° 43' 32". 

.-. F = 119° 41' 30", S = 55° 45' 34". 

log sin N = 9.99913- 10 

log sin s = 9.89772 - 10 

.-. log product = 9.89685 - 10 

log sin S = 9.91734 - 10 

.-. log sin n= 9.97951 

.-. n = 107° 27' 45". 

SinceZF=119°4r 30", n + s>/+l°. 

Hence when leaving San Francisco the ship's course is S. 60° 18' 30" W., 
and when approaching Sydney it is S. 55° 45' 34" W. The distance sailed 
is 107.46^° = 107.46J X 22 it mi. = 7,427.3 mi. 

9. In example 8 find the distance sailed and the longitude and the course 
of the ship when it crosses the equator. 

In A EFN let the vertex F be San Francisco, N the North Pole, and 
E the point where the great circle through San Francisco and Sydney 
cuts the equator; then 

e = 52° 12' 5", /= 90°, F= 119° 41' 30". 



EXERCISE XLVI 



203 



To find N, n, and E, we have the formulas 
sin E = sin e sin F/sin /, 
tan (n/2) = sin | (F + £) tan \ (f - e)/sin \(F -E), 
cot (JV/2) = sin | (/ + e) tan | (F - E)/sin \ (f - e). 



log sin e= 9.89772 - 10 
log sin F = 9.93887 - 10 



log product = 9.83659 
log sin / = 0.00000 



10 



.-. log sin # = 9.83659 - 10 

.-. E = 43° 20' 51". 

E < F, since e < f. 

log sin l(F + E) = 9.99522 - 10 

log tan i (/ - e) = 9.53449 - 10 

10 

10 



10 



.-. log product = 9.52971 
log sin i (F. - E) == 9.79100 
.-. log tan (n/2) = 9.73871 

.-. n/2 = 28° 43' 8". 
.-. n = 57° 26' 16". 
Hence the longitude of E is (122° 24' 32" -} 
the course of the ship at E is S. 43° 20' 51' 
is 57.44° = 57.44 x 22 n mi. = 3969.8 mi. 



(F + E)/2 = 81° 31' 10". 
(F - -E)/2 = 38° 10' 20". 

(/ - e)/2 = 18° 53' 58". 

(/+e)/2 = 71°6'3". 



log sin I (/ + e) = 9. 97593 - 10 
log tan I \f - E) = 9.89550 - 10 
.-. log product = 9.87143 - 10 
log sin i (/ - e) = 9.51042 - 10 
.-. log cot (N/2) = 0.36101 
.-. N/2 = 23° 32'. 



.-. N = 47° 4'. 
47° 4') W. , or 169° 28' 32" W. ; 
W. ; and the distance sailed 



EXERCISE XLVI 

1. At a place in latitude 42° N. the altitude of a star whose declina- 
tion is + 60° was measured and found to be 50°, the star being east of 
the meridian. How many hours 
afterward did the star cross the 
meridian ? 

Here we have given a = ZS = 40°, 
b = ZP = 48°, c = PS = 30°; to find 
(Z ZPS/15 ), or (^4/15°) hr. 





A 

~2 ' 


■V 


n(s - 


- b) sin (s — 


c) 




sin s sin (s — a) 






s - 


= 59°, 








s - 


- a - 


= 19°, 








s - 


-6 = 


= 11°, 








s - 


- c = 


= 29°. 










Fig. 17 



204 SPHERICAL TRIGONOMETRY 

log sin (s - b) = 9.28060 - 10 

log sin {s-c) = 9.68557 - 10 

.-. log numerator = 8.96617 — 10 

log sin s= 9.93307 -10 

log sin s - a = 9.51264 -1 

•. log denominator = 9.44571 —10 

.-. log tan (A/2) = (9.52046 - 10)/2. 

= 9.76023-10. 

.-. A = (29° 55' 56") • 2 

= 59° 51' 52" = 59° 51. 9'. 

59° 51 9' 

.-. Time = — hr. = 3 hr. 59.5 min. 

15° 

2. The declination of a star is + 10°; find the length of time between 
its rising above the horizon and its passage across the meridian, the place 
of observation being in latitude 42° N. 

Given ZS = 90°, ZP = 48°, PS = 80° ; to find (Z ZPS/15 ) hr. 

cos ZS = cos ZP cos PS + sin ZP sin PS cos ZPS. 
Here cos ZS = 0. 

.-. cos ZPS = - cot ZP cot PS. 
log cot ZP = 9.95444 - 10 
log cot PS = 9.24632 - 10 
.-. log cos ZPS = 9.20076 - 10 

.-. ZZPS = 99° 8.1'. 

99° 8 V 
... Time = hr. = 6 hr. 36.5 min. 

15° 

3. The declination of a star is + 20° ; find the interval between the 
instant when it is due east and when it is due west, the place of observa- 
tion being in latitude 42° N. 

Here A PZS = 90°, ZP = 90° - 42° = 48°, SP = 90° - 20° = 70° ; to 

find (2 Z ZPS/ 15°) hr. 

cos ZPS = tan ZP cot SP = tan 48° cot 70°. 

log tan 48° = 0.04556 
log cot 70° = 9.56107 - 10 
.-. log cos ZPS = 9.60663 - 10 
.-. Z ZPS = 66° 9. 4'. 
.-. 2ZZP£= 132° 18.8'. 

... Time = 182 ° 18 ' 8 - hr. = 8 hr. 49.3 min. 
15° 



EXERCISE XLVI 205 

4. In New York, 40° 43' N., at a forenoon observation the sun's alti- 
tude is 30° 40'. Given that the sun's declination is + 10°, find the time 
of day of the observation. 

At noon the sun is on the meridian PZH ; hence the time required is 
(Z ZPS/15 ) hr. before noon. 

Given a = SZ = 90° - 30° 40' = 59° 20', b = PZ = 90° - 40° 43' = 49° 17', 
c = SP = 90° - 10° = 80° ; to find (Z ZP£/15°), or (A/15°) hr. 

J. /sin (s — b) sin (s — c) 



tan 



Vsm(s 
sir 



2 \ sin s sin (s — a) 



log sin (s - 6) = 9.39295 - 10 s - 6 = 14° 18' 30". 

log sin (s-c)= 9.84967 -10 s - c = 45° V 30". 

.-. log numerator = 19.24262 - 20 s = 94° 18' 30". 

log sin s = 9.99878 - 10 s - a = 34° 58' 30". 

log sin (s-a)= 9.75832 -10 
.-. log tan (A/2) = (19.48552 - 20)/2 = 9.74276 - 10. 

.-. ^4/2 = 28° 56 7 40". 
.-. Z ZPS = 57° 53' 20". 

(57° 53' 20"/15°) hr. before noon = 3 hr. 51.6 min. before noon, or 
8 :8.4 a.m. 

5. In Montreal, 45° 30' N., at an afternoon observation the sun's alti- 
tude is 26° 30 7 . Given that the sun's declination is 8°. S., find the time of 
day of the observation. 

Here a = SZ = 63° 30', b = PZ = 44° 30', c = SP = 98°, and the time 
required is (Z ZP£/15°), or (-4/15°) hr. after noon. 

A /sin (s — b) sin (s — c) 



tan — = 

2 \ sin s sin (s — a) 




log sin (s - b) = 8.94030 - 10 




s = 103°. 


log sin (s -c)= 9. 93077 - 10 




s - a = 39° 30'. 


log numerator = 18.87107 —20 




s - b = 5°. 


log sin s= 9.98872-10 




s - c = 58° 30 7 . 


log sin (s- a)= 9.80351 - 10 






log tan (A/2) = (9.07884 - 10)/2 = 


= 9.53942 - 


-10. 


.-. A/2 = 19° 5' 59". 






.-. Z ZP-S = 38° 11' 57". 







(38° 11' 57"/15°) hr. after noon = 2 hr. 33 min. after noon, or 2 : 33 p.m. 



206 SPHERICAL TRIGONOMETRY 

6. In New York, 40° 43' N. , at a forenoon observation the sun's altitude 
is 30° 40'. Given that the sun's declination is 10° S., find the time of day 
of the observation. 

Here a = ZS = 59° 20', b = ZP = 49° 17', c = SP = 100° ; to find 
(Z ZP£/15°), or (4/15°) hr. 

4 _ 'sin (s — b) sin (s — c) 

tan — — ^ / • 

2 \ sin s sin (s — a) 

log sin (s-b)= 8.87578 - 10 s = 104° 18' 30". 

log sin (s - c) = 9.91349-10 s - a = 44° 58' 30". 

.-. log numerator = 18.78927 - 20 s - b = 4° 18' 30". 

log sin s = 9. 98632 -10 s - c = 55° 1' 30". 
log sin (s - a) = 9.84929-10 
.-. log tan (4/2) = (18.95366)/2 = 9.47683 - 10. 

.-. 4/2 = 16° 4r 20". 

.-. A = r Z ZPS = 33° 22' 39". 

(33° 22' 39"/15°) hr. before noon = 2 hr. 13.5 min. before noon, or 9 : 46.5 a.m. 

7. In Montreal, 45° 30' N., at an afternoon observation the sun's alti- 
tude is 26° 30'. Given that the sun's declination is 8° S. , find the time of 
day of the observation. 

Here a = SZ = 63° 30', b = PZ = 44° 30', c = SP = 72°, and the time 
required is (ZZPS/15 ), or (4/15°) hr., after noon. 

/sin (s — b) sin (s — c) 



A _ k 
2"\ 



tan 

sin s sin (s — a) 



log sin (s - b) = 9.48998 - 10 s = 90°. 

log sin (s - c) = 9.85324 - 10 s - a = 26° 30 r . 

.-. log numerator = 19.34322 - 20 s - b = 18°. 

log sin s= 0.00000 s - c; = 45° 30'. 

log sin (s - a) = 9.64953 - 10 
.-. log tan (4/2)= (9.69369 - 10)/2 = 9.84684 - 10. 

.-•. 4/2 = 35° 6'. 
.-. Z ZPS = 70° 12'. 

(Z ZPS/1& ) hr. after noon = 4 hr. 40.8 min. after noon, or 4:40.8 p.m. 

8. Find the approximate time of sunrise in New York, 40°43 / N., 
about April 1, when the declination of the sun is 4° 30' N. 

Here ZP = 49° 17', SP = 85° 30', SZ = 90°; to find \Z ZPS/lb ) hr. 
before noon. 



EXERCISE XLVI 207 

By example 2, cos ZP-S = — cot ZP cot -SP. 

log cot ZP = 9.93482 - 10 
log cot -SP = 8.89598 -10 
.-. log cos ZPS = 8.83080 - 10 
.-. ZPS = 93° 53' 2 // . 
.-. (Z ZP-S/15 ) hr. before noon = 6 hr. 15 min. 32 sec. before noon, or 
5:44.5 a.m. 

9. Find the approximate time of sunrise in London, 51° 30' 48" N,, 
about October 8, when the declination of the sun is 6° S. 

Here ZP = 38° 29' 12", -SP = 96°, SZ = 90°; to find (Z ZP-S/15 ) hr. 
before noon. 

By example 2, cos ZP-S = - cot ZP cot -SP. 

log cot ZP= 0.09960 
log cot -SP = 9.02162 - 10 
.-. log cos ZP-S = 9.12122 - 10 
.-. Z ZPS = 82° 24' 13". 
.-. (ZZP-S/15 ) hr. before noon = 5 hr. 29 min. 37 sec. before noon, or 
6:30.4 a.m. 

10. Find the approximate time of sunrise in New Orleans, 29° 58' N., 
about September 10, when the declination of the sun is 5° N. 

Here ZP=60°2', -SP=85°, -SZ = 90°; to find (Z ZP-S/15°) hr. before noon. 

By example 2, cos ZPS - - cot ZP cot -SP 

= - cot 60° 2' • cot 85°. 

log cot 60° 2' = 9.76086 - 10 

log cot 85° = 8.94195-10 

.-. cos ZP-S = 8.70281 -10 

.-. Z ZPS =90° + 2° 53. 5'. 

92° 53 5' 
Time of sunrise is — — hr. before noon, or 6 hr. 11.6 min. before 

noon, or 5:48.4 a.m. 

11. Given the latitude of the place 52° 30' 16", the declination of a star 
38°, and its hour angle 331° 42' 45"; find the altitude. 

Here a = -SP = 52°, b=ZP= 37° 29' 44", C = Z ZPS = 28° 17' 15"; to 
find 90° - c, or h. 

tani(J. + B) = cosi(a -b)cot(C/2)/cosi(a + b). 
tan i (A - B) = sin £ (a - b) cot (C/2)/sin i (a + 6). 
sin c = sin a sin O/sin A. 
(a + 6)/2 = 44° 44' 52". 
(a - 6)/2 = 7° 15' 8". 
C/2 = 14° 8' 37.5". 



208 



SPHERICAL TRIGONOMETRY 



log cos £ (a - b) = 9. 99651 - 1 
log cot ( C/2) = 0.59861 
.-. log product = 10.59512 - 10 
log cos £ (a + 5)= 9.85139-10 
•.logtan£(^. + £)= 0.74373 

.-. (A + B)/2 = 79° 46' 23". 
.-. A = 115° 12' 30", 
B = 44° 20' 16". 



log sin £ (a — b) 

log cot (C/2) 

.-. log product 

log sin £ (a -f b) 

. log tan i (A- B) 

.-. (A - B)/2 

log sin a 

log sin 6 

.-. log product 

log sin A 

.-. log sine 



9.10119-10 
0.59861 



19.69980 
9.84756 



20 
10 



•. h = 90 c 



= 9.85224 - 10 

= 35° 26' 7". 

= 9.89653 - 10 
= 9.67568-10 
= 19.57221 -20 
= 9.95654 - 10 
= 9.61567 - 10 

= 24° 22' 36". 
= 65° 37' 24". 



12. Given the latitude of the place 51° 19' 20", the declination of a star 
22° 0' 55", and its hour angle 344° 51' 48"; find its altitude and its azimuth. 



Here a = SP = 67° 59' 5", b = ZP = 38° 40' 40' 
15° 8' 12"; to find SZ and 180° + Z SZP. 



C=Z ZPS = 360° - * 



tan \ (A + B) = cos £ (a - b) cot (C/2)/cos i (a + b). 
tan£(J. - B) = sin£(a - &)cot(C/2)/sin£(a + b). 
sin c = sin a sin C/sin A. 
(a + 6)/2 = 53° 19' 53". 
(a - 6)/2 = 14° 39' 13". 
C/2 = 7° 34' 6". 



log 


cos £ (a — b) = 


9.98564 - 


10 


log sin £ (a — b) = 


9.40308 - 


-10 




log cot C/2 = 


0.87658 




log cot ( C/2) = 


0.87658 






log product = 


10.86222 - 


10 


.-. log product = 


10.27966 - 


-10 


log 


cos i (a + b) = 


9.77611 - 


10 


log sin £ (a + b) = 


9.90423 - 


-10 


.log 


tan£(^i + 5) = 


1.08611 




.:\ogta,ni(A-B) = 


0.37543 






. (A + B)/2 = 


85° 18' 41" 




.-. (A - B)/2 = 


67° 9' 19". 






.-. A = 


152° 28', 




log sin a = 


9.96711 - 


-10 




B = 


18° 9' 22". 




log sin C = 

.-. log product = 

log sin A = 

.-. log sine = 


9.41684 - 

19.38395 - 

9.66489 - 


-10 
-20 
-10 




9.71906 - 


-10 










.-. c = 


31° 34' 45' 








ft = 90° 


-31° 


1 34' 45" = 58° 25' 15" 








Azimuth = 152 c 


3 28'H 


h 180° = 332° 28'. 







EXERCISE XL VI 



209 



13. Given the declination of a star 7° 54', its altitude 22° 45' 12", and 
its azimuth 50° 14' 23"; find its hour angle and the latitude of the observer. 



Here a 
50° 14' 23' 



SP = 82° 6', 
= 129° 45' 37": 



6 = SZ = 07° 14' 48", J. = Z SZP = 180° - 
to find Z ZPS and ZP, or JB and c. 



sin B = sin 6 sin A /sin a. 
tan(c/2) = sini(^L + 7?) tan i (a - b)/sm$(A- B). 



log sin 6= 9.96482-10 
log sin A = 9.88577 - 10 

log product = 19.85059 - 20 
log sin a = 9.99586 -10 

.-. logsinJ5= 9.85473 - 10 

.-. t=B = 45° 42'. 



Hence /3 = 90° 



67° 58' 56' 



(a + 6)/2 = 74° 40' 24" 

(a - 6)/2 = 7° 25' 36". 

(^4 + B)/2 = 87° 43' 49" 

(A -B)/2 = 42° 1' 49". 

log sin $(A + B) = 9. 99966 - 



10 



log tan i (a - b) = 9.11511 -10 

.-. log product = 19.11477 -20 

log sin i (A -B) = 9.82576 - 10 

.-. log tan (c/2) = 9.28901-10 

.-. c/2 = 11° 0' 32", or c = 22° Y 4" 



14. Given the latitude of the place 44° 50' 14", the azimuth of a star 
138° 58' 43", and its hour angle 20° ; find its declination and its altitude. 



Here A 

45° 9' 46": 



= Z SZP = 41° Y 17", B = Z ZPS 
to find -SP and SZ, or a and b. 



20 c 



ZP = 90 c 



tan £ (a + b) = cos £ (A — P) tan $ c/cos £ (^4 -f- 5). 
tan i(a — 6) = sin £ ( J. — B) tan £ c/sin i (^1 + B). 

(A + B)/2 = 30° 30' 39". 

(A - B)/2 = 10° 30' 39". 
c/2 = 22° 34' 53". 



log cos i(A-B)= 9. 99265 - 10 
log tan (c/2) = 9.61897 - 10 


= 16 
73 c 


log sin £(4 -£) = 9.26107 - 

log tan (c/2) = 9.61897 - 

.-. log product = 18.88004- 

log sin i(A + B)= 9. 70561 - 

.-. logtani(a-6)= 9.17443- 

.-. (a - b)/2 = 8° 29' 55" 
° 53' 29". 
6' 31". 


-10 
-10 


.-. log product = 19.61162 - 20 
logcosi(^l + P)= 9.93527-10 


-20 
-10 


.-. log tan i (a + &) = 9.67635 - 10 

.-. (a + 6)/2 = 25° 23' 24". 

.-. a = 33° 53' 19", b = 
.-. 5 = 56° 6' 41", /i = 


-10 



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THIS text-book aims to present clearly, scientifically, and in their 
true relations the three common methods in the Calculus. 
The concept of rates gives a clear idea and statement of the 
problems of the Calculus ; the principles of limits afford general solutions 
of these problems ; and the laws of infinitesimals greatly abridge these 
solutions. 

The method of rates is so generalized and simplified that it does not 
involve " the foreign element of time," and affords the simplest and 
briefest proofs of first principles. 

By proving It (Ay /Ax) = dy/dx, the problem of rates is reduced to 
one of limits or infinitesimals. 

The theory of infinitesimals is viewed simply as that part of the theory 
of limits which deals with variables having zei-o as their common limit. 

The important distinction between infinitesimals and zero and that 
between infinites and a/o are emphasized. 

Taylor's theorem is accurately stated and proved. 

The method of finding asymptotes illustrates the meaning of impos- 
sible and defective systems of equations. 

The applications of double and triple integration clearly set forth the 
meanings of these operations. 

A chapter on differential equations is added to meet an increasing 
demand for a short course in this subject. 

A table of integrals, for convenience of reference, is appended. 

Throughout the work the numerous problems so set forth and illustrate 
the highly practical nature of the Calculus as to awaken in the reader a 
lively interest. 

Those who prefer to study the Calculus by the method of limits alone 
can omit the few demonstrations which involve rates and take in their 
stead those by limits or infinitesimals. 



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